Answer on Question #51421 – Math – Integral Calculus:
∫ d x x 3 + x 8 = ? \int \frac{dx}{x^3 + x^8} = ? ∫ x 3 + x 8 d x = ?
Solution
Denote ϕ = 5 − 1 2 \phi = \frac{\sqrt{5} - 1}{2} ϕ = 2 5 − 1
ϕ 2 = 6 − 2 5 4 = 3 − 5 2 = 1 − ϕ ; \phi^2 = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} = 1 - \phi; ϕ 2 = 4 6 − 2 5 = 2 3 − 5 = 1 − ϕ ;
We have that:
( x 2 + ϕ x + 1 ) ( x 2 − 1 ϕ x + 1 ) = x 4 + ( ϕ − 1 ϕ ) x 3 + x 2 + ( ϕ − 1 ϕ ) x + 1 = x 4 + ϕ 2 − 1 ϕ x 3 + x 2 + ϕ 2 − 1 ϕ x + 1 = x 4 − x 3 + x 2 − x + 1. \begin{aligned}
(x^2 + \phi x + 1)(x^2 - \frac{1}{\phi} x + 1) &= x^4 + \left(\phi - \frac{1}{\phi}\right) x^3 + x^2 + \left(\phi - \frac{1}{\phi}\right) x + 1 \\
&= x^4 + \frac{\phi^2 - 1}{\phi} x^3 + x^2 + \frac{\phi^2 - 1}{\phi} x + 1 \\
&= x^4 - x^3 + x^2 - x + 1.
\end{aligned} ( x 2 + ϕ x + 1 ) ( x 2 − ϕ 1 x + 1 ) = x 4 + ( ϕ − ϕ 1 ) x 3 + x 2 + ( ϕ − ϕ 1 ) x + 1 = x 4 + ϕ ϕ 2 − 1 x 3 + x 2 + ϕ ϕ 2 − 1 x + 1 = x 4 − x 3 + x 2 − x + 1.
So:
x 3 + x 8 = x 3 ( x 5 + 1 ) = x 3 ( x + 1 ) ( x 4 − x 3 + x 2 − x + 1 ) = x 3 ( x + 1 ) ( x 2 + ϕ x + 1 ) ( x 2 − 1 ϕ x + 1 ) ; \begin{aligned}
x^3 + x^8 &= x^3(x^5 + 1) = x^3(x + 1)(x^4 - x^3 + x^2 - x + 1) \\
&= x^3(x + 1)(x^2 + \phi x + 1)\left(x^2 - \frac{1}{\phi} x + 1\right);
\end{aligned} x 3 + x 8 = x 3 ( x 5 + 1 ) = x 3 ( x + 1 ) ( x 4 − x 3 + x 2 − x + 1 ) = x 3 ( x + 1 ) ( x 2 + ϕ x + 1 ) ( x 2 − ϕ 1 x + 1 ) ;
Expand 1 x 3 + x 8 \frac{1}{x^3 + x^8} x 3 + x 8 1
1 x 3 + x 8 = a x + b x 2 + c x 3 + d x + 1 + f x + g x 2 + ϕ x + 1 + h x + k x 2 − 1 ϕ x + 1 ; \frac{1}{x^3 + x^8} = \frac{a}{x} + \frac{b}{x^2} + \frac{c}{x^3} + \frac{d}{x + 1} + \frac{fx + g}{x^2 + \phi x + 1} + \frac{hx + k}{x^2 - \frac{1}{\phi} x + 1}; x 3 + x 8 1 = x a + x 2 b + x 3 c + x + 1 d + x 2 + ϕ x + 1 f x + g + x 2 − ϕ 1 x + 1 h x + k ;
Now find a , b , c , d , f , g , h , k a, b, c, d, f, g, h, k a , b , c , d , f , g , h , k
1 = a x 2 ( x 5 + 1 ) + b x ( x 5 + 1 ) + c ( x 5 + 1 ) + d x 3 ( x 4 − x 3 + x 2 − x + 1 ) + + ( f x + g ) x 3 ( x + 1 ) ( x 2 − 1 ϕ x + 1 ) + ( h x + k ) x 3 ( x + 1 ) ( x 2 + ϕ x + 1 ) ⇒ ⇒ 1 = ( a + d + f + h ) x 7 + ( b − d + g + f − f ϕ + h + k + ϕ h ) x 6 + + ( c + d + g − g ϕ − f ϕ + f + k + ϕ k + ϕ h + h ) x 5 + + ( − d + f + g − g ϕ + h + k + ϕ k ) x 4 + ( d + g + k ) x 3 + a x 2 + b x + c ⇒ \begin{aligned}
1 &= a x^2(x^5 + 1) + b x(x^5 + 1) + c(x^5 + 1) + d x^3(x^4 - x^3 + x^2 - x + 1) + \\
&+ (f x + g) x^3(x + 1)\left(x^2 - \frac{1}{\phi} x + 1\right) + (h x + k) x^3(x + 1)(x^2 + \phi x + 1) \Rightarrow \\
&\Rightarrow 1 = (a + d + f + h) x^7 + \left(b - d + g + f - \frac{f}{\phi} + h + k + \phi h\right) x^6 + \\
&\quad + \left(c + d + g - \frac{g}{\phi} - \frac{f}{\phi} + f + k + \phi k + \phi h + h\right) x^5 + \\
&\quad + \left(- d + f + g - \frac{g}{\phi} + h + k + \phi k\right) x^4 + (d + g + k) x^3 + a x^2 + b x + c \Rightarrow
\end{aligned} 1 = a x 2 ( x 5 + 1 ) + b x ( x 5 + 1 ) + c ( x 5 + 1 ) + d x 3 ( x 4 − x 3 + x 2 − x + 1 ) + + ( f x + g ) x 3 ( x + 1 ) ( x 2 − ϕ 1 x + 1 ) + ( h x + k ) x 3 ( x + 1 ) ( x 2 + ϕ x + 1 ) ⇒ ⇒ 1 = ( a + d + f + h ) x 7 + ( b − d + g + f − ϕ f + h + k + ϕ h ) x 6 + + ( c + d + g − ϕ g − ϕ f + f + k + ϕ k + ϕ h + h ) x 5 + + ( − d + f + g − ϕ g + h + k + ϕ k ) x 4 + ( d + g + k ) x 3 + a x 2 + b x + c ⇒ ⇒ { a = b = 0 , c = 1 a + d + f + h = 0 d + g + k = 0 − d + f + g − g ϕ + h + k + ϕ k = 0 c + d + g − g ϕ − f ϕ + f + k + ϕ k + ϕ h + h = 0 b − d + g + f − f ϕ + h + k + ϕ h = 0 ⇒ { a = b = 0 , c = 1 d + f + h = 0 d + g + k = 0 − 3 d − g ϕ + ϕ k = 0 1 − d − g ϕ − f ϕ + ϕ k + ϕ h = 0 − 3 d − f ϕ + ϕ h = 0 ⇒ { a = b = 0 , c = 1 d + f + h = 0 g + k = f + h − g ϕ + ϕ k = 3 d 1 − d − g ϕ − f ϕ + ϕ k + ϕ h = 0 ϕ ( h − k ) = f − g ϕ ⇒ { a = b = 0 , c = 1 d + f + h = 0 g + k = f + h − g ϕ + ϕ k = 3 d 1 − d − g ϕ − f ϕ + ϕ k + ϕ h = 0 ( f − g ) ( ϕ + 1 ϕ ) = 0 ⇒ { a = b = 0 , c = 1 d + f + h = 0 g + k = f + h − g ϕ + ϕ k = 3 d 1 − d − g ϕ − f ϕ + ϕ k + ϕ h = 0 ( f − g ) = 0 ⇒ { a = b = 0 , c = 1 d + f + h = 0 h = k − f ϕ + ϕ h = 3 d 1 − d − 2 f ϕ + 2 ϕ h = 0 f = g ⇒ { a = b = 0 , c = 1 d + f + h = 0 h = k ϕ h − 3 d = f ϕ 1 + 5 d = 0 f = g ⇒ { a = b = 0 , c = 1 f + h = 1 5 h = k ϕ h + 3 5 = f ϕ d = − 1 5 f = g \begin{array}{l}
\Rightarrow \left\{
\begin{array}{c}
a = b = 0, c = 1 \\
a + d + f + h = 0 \\
d + g + k = 0 \\
-d + f + g - \frac{g}{\phi} + h + k + \phi k = 0 \\
c + d + g - \frac{g}{\phi} - \frac{f}{\phi} + f + k + \phi k + \phi h + h = 0 \\
b - d + g + f - \frac{f}{\phi} + h + k + \phi h = 0
\end{array}
\right. \\
\Rightarrow \left\{
\begin{array}{c}
a = b = 0, c = 1 \\
d + f + h = 0 \\
d + g + k = 0 \\
-3d - \frac{g}{\phi} + \phi k = 0 \\
1 - d - \frac{g}{\phi} - \frac{f}{\phi} + \phi k + \phi h = 0 \\
-3d - \frac{f}{\phi} + \phi h = 0
\end{array}
\right.
\Rightarrow \left\{
\begin{array}{c}
a = b = 0, c = 1 \\
d + f + h = 0 \\
g + k = f + h \\
-\frac{g}{\phi} + \phi k = 3d \\
1 - d - \frac{g}{\phi} - \frac{f}{\phi} + \phi k + \phi h = 0 \\
\phi(h - k) = \frac{f - g}{\phi}
\end{array}
\right. \\
\Rightarrow \left\{
\begin{array}{c}
a = b = 0, c = 1 \\
d + f + h = 0 \\
g + k = f + h \\
-\frac{g}{\phi} + \phi k = 3d \\
1 - d - \frac{g}{\phi} - \frac{f}{\phi} + \phi k + \phi h = 0 \\
(f - g)(\phi + \frac{1}{\phi}) = 0
\end{array}
\right.
\Rightarrow \left\{
\begin{array}{c}
a = b = 0, c = 1 \\
d + f + h = 0 \\
g + k = f + h \\
-\frac{g}{\phi} + \phi k = 3d \\
1 - d - \frac{g}{\phi} - \frac{f}{\phi} + \phi k + \phi h = 0
\end{array}
\right. \\
( f - g ) = 0
\end{array}
\Rightarrow \left\{
\begin{array}{c}
a = b = 0, c = 1 \\
d + f + h = 0 \\
h = k \\
-\frac{f}{\phi} + \phi h = 3d \\
1 - d - \frac{2f}{\phi} + 2\phi h = 0 \\
f = g
\end{array}
\right.
\Rightarrow \left\{
\begin{array}{c}
a = b = 0, c = 1 \\
d + f + h = 0 \\
h = k \\
\phi h - 3d = \frac{f}{\phi} \\
1 + 5d = 0 \\
f = g
\end{array}
\right.
\Rightarrow \left\{
\begin{array}{c}
a = b = 0, c = 1 \\
f + h = \frac{1}{5} \\
h = k \\
\phi h + \frac{3}{5} = \frac{f}{\phi} \\
d = -\frac{1}{5} \\
f = g
\end{array}
\right. ⇒ ⎩ ⎨ ⎧ a = b = 0 , c = 1 a + d + f + h = 0 d + g + k = 0 − d + f + g − ϕ g + h + k + ϕ k = 0 c + d + g − ϕ g − ϕ f + f + k + ϕ k + ϕ h + h = 0 b − d + g + f − ϕ f + h + k + ϕ h = 0 ⇒ ⎩ ⎨ ⎧ a = b = 0 , c = 1 d + f + h = 0 d + g + k = 0 − 3 d − ϕ g + ϕ k = 0 1 − d − ϕ g − ϕ f + ϕ k + ϕ h = 0 − 3 d − ϕ f + ϕ h = 0 ⇒ ⎩ ⎨ ⎧ a = b = 0 , c = 1 d + f + h = 0 g + k = f + h − ϕ g + ϕ k = 3 d 1 − d − ϕ g − ϕ f + ϕ k + ϕ h = 0 ϕ ( h − k ) = ϕ f − g ⇒ ⎩ ⎨ ⎧ a = b = 0 , c = 1 d + f + h = 0 g + k = f + h − ϕ g + ϕ k = 3 d 1 − d − ϕ g − ϕ f + ϕ k + ϕ h = 0 ( f − g ) ( ϕ + ϕ 1 ) = 0 ⇒ ⎩ ⎨ ⎧ a = b = 0 , c = 1 d + f + h = 0 g + k = f + h − ϕ g + ϕ k = 3 d 1 − d − ϕ g − ϕ f + ϕ k + ϕ h = 0 ( f − g ) = 0 ⇒ ⎩ ⎨ ⎧ a = b = 0 , c = 1 d + f + h = 0 h = k − ϕ f + ϕ h = 3 d 1 − d − ϕ 2 f + 2 ϕ h = 0 f = g ⇒ ⎩ ⎨ ⎧ a = b = 0 , c = 1 d + f + h = 0 h = k ϕ h − 3 d = ϕ f 1 + 5 d = 0 f = g ⇒ ⎩ ⎨ ⎧ a = b = 0 , c = 1 f + h = 5 1 h = k ϕ h + 5 3 = ϕ f d = − 5 1 f = g \begin{array}{l}
\Rightarrow \left\{
\begin{array}{c}
a = b = 0, c = 1 \\
h = \frac{1}{5} - f \\
h = k \\
\phi\left(\frac{1}{5} - f\right) + \frac{3}{5} = \frac{f}{\phi} \\
d = -\frac{1}{5} \\
f = g
\end{array}
\right.
\Rightarrow
\left\{
\begin{array}{c}
a = b = 0, c = 1 \\
h = \frac{1}{\phi} - 3 \\
5\left(\phi + \frac{1}{\phi}\right) \\
k = -\frac{1}{\phi} - 3 \\
5\left(\phi + \frac{1}{\phi}\right) \\
f = \frac{\phi + 3}{5\left(\phi + \frac{1}{\phi}\right) \\
d = -\frac{1}{5} \\
g = \frac{\phi + 3}{5\left(\phi + \frac{1}{\phi}\right)}
\end{array}
\right.
\Rightarrow \\
\Rightarrow \left\{
\begin{array}{c}
a = b = 0, c = 1 \\
d = -\frac{1}{5} \\
f = g = \frac{1}{5\phi} \\
h = k = -\frac{\phi}{5}
\end{array}
\right.
\end{array}
So:
∫ d x x 3 + x 8 = ∫ ( 1 x 3 − 1 5 ( x + 1 ) + 1 5 ϕ ⋅ x + 1 x 2 + ϕ x + 1 − ϕ 5 ⋅ x + 1 x 2 − 1 ϕ x + 1 ) d x = = ∫ d x x 3 − 1 5 ∫ d x x + 1 + 1 5 ϕ ∫ x + 1 x 2 + ϕ x + 1 d x − ϕ 5 ∫ x + 1 x 2 − 1 ϕ x + 1 d x = = − 1 2 x 2 − 1 5 ln ∣ x + 1 ∣ + 1 5 ϕ ∫ x + 1 x 2 + ϕ x + 1 d x − ϕ 5 ∫ x + 1 x 2 − 1 ϕ x + 1 d x ; \begin{array}{l}
\int \frac{dx}{x^3 + x^8} = \int \left( \frac{1}{x^3} - \frac{1}{5(x + 1)} + \frac{1}{5\phi} \cdot \frac{x + 1}{x^2 + \phi x + 1} - \frac{\phi}{5} \cdot \frac{x + 1}{x^2 - \frac{1}{\phi} x + 1} \right) dx = \\
= \int \frac{dx}{x^3} - \frac{1}{5} \int \frac{dx}{x + 1} + \frac{1}{5\phi} \int \frac{x + 1}{x^2 + \phi x + 1} dx - \frac{\phi}{5} \int \frac{x + 1}{x^2 - \frac{1}{\phi} x + 1} dx = \\
= -\frac{1}{2x^2} - \frac{1}{5} \ln |x + 1| + \frac{1}{5\phi} \int \frac{x + 1}{x^2 + \phi x + 1} dx - \frac{\phi}{5} \int \frac{x + 1}{x^2 - \frac{1}{\phi} x + 1} dx;
\end{array} ∫ x 3 + x 8 d x = ∫ ( x 3 1 − 5 ( x + 1 ) 1 + 5 ϕ 1 ⋅ x 2 + ϕ x + 1 x + 1 − 5 ϕ ⋅ x 2 − ϕ 1 x + 1 x + 1 ) d x = = ∫ x 3 d x − 5 1 ∫ x + 1 d x + 5 ϕ 1 ∫ x 2 + ϕ x + 1 x + 1 d x − 5 ϕ ∫ x 2 − ϕ 1 x + 1 x + 1 d x = = − 2 x 2 1 − 5 1 ln ∣ x + 1∣ + 5 ϕ 1 ∫ x 2 + ϕ x + 1 x + 1 d x − 5 ϕ ∫ x 2 − ϕ 1 x + 1 x + 1 d x ;
Compute the rest of integrals:
∫ x + 1 x 2 + ϕ x + 1 d x = ∫ x + 1 ( x + ϕ 2 ) 2 + 1 − ϕ 2 4 d x = [ y = x + ϕ 2 1 − ϕ 2 4 ] = = ∫ 1 − ϕ 2 4 y + 1 − ϕ 2 ( 1 − ϕ 2 4 ) ( y 2 + 1 ) ⋅ 1 − ϕ 2 4 d y = ∫ y + 1 − ϕ 2 1 − ϕ 2 4 y 2 + 1 d y = \begin{array}{l}
\int \frac{x + 1}{x^2 + \phi x + 1} dx = \int \frac{x + 1}{\left(x + \frac{\phi}{2}\right)^2 + 1 - \frac{\phi^2}{4}} dx = \left[ y = \frac{x + \frac{\phi}{2}}{\sqrt{1 - \frac{\phi^2}{4}}} \right] = \\
= \int \frac{\sqrt{1 - \frac{\phi^2}{4}} y + 1 - \frac{\phi}{2}}{\left(1 - \frac{\phi^2}{4}\right) (y^2 + 1)} \cdot \sqrt{1 - \frac{\phi^2}{4}} dy = \int \frac{y + \frac{1 - \frac{\phi}{2}}{\sqrt{1 - \frac{\phi^2}{4}}}}{y^2 + 1} dy = \\
\end{array} ∫ x 2 + ϕ x + 1 x + 1 d x = ∫ ( x + 2 ϕ ) 2 + 1 − 4 ϕ 2 x + 1 d x = [ y = 1 − 4 ϕ 2 x + 2 ϕ ] = = ∫ ( 1 − 4 ϕ 2 ) ( y 2 + 1 ) 1 − 4 ϕ 2 y + 1 − 2 ϕ ⋅ 1 − 4 ϕ 2 d y = ∫ y 2 + 1 y + 1 − 4 ϕ 2 1 − 2 ϕ d y = = ∫ y y 2 + 1 d y + 1 − ϕ 2 1 − ϕ 2 4 ∫ d y y 2 + 1 = 1 2 ∫ d ( y 2 ) y 2 + 1 + 1 − ϕ 2 1 − ϕ 2 4 arctan y = = 1 2 ln ( y 2 + 1 ) + 1 − ϕ 2 1 − ϕ 2 4 arctan y + c = = 1 2 ln ( ( x + ϕ 2 1 − ϕ 2 4 ) 2 + 1 ) + 1 − ϕ 2 1 − ϕ 2 4 arctan x + ϕ 2 1 − ϕ 2 4 + c = = 1 2 ln ( x 2 + ϕ x + 1 1 − ϕ 2 4 ) + 1 − ϕ 2 1 − ϕ 2 4 arctan x + ϕ 2 1 − ϕ 2 4 + c ; \begin{array}{l}
= \int \frac{y}{y^{2} + 1} \, dy + \frac{1 - \frac{\phi}{2}}{\sqrt{1 - \frac{\phi^{2}}{4}}} \int \frac{dy}{y^{2} + 1} = \frac{1}{2} \int \frac{d(y^{2})}{y^{2} + 1} + \frac{1 - \frac{\phi}{2}}{\sqrt{1 - \frac{\phi^{2}}{4}}} \arctan y = \\
= \frac{1}{2} \ln(y^{2} + 1) + \frac{1 - \frac{\phi}{2}}{\sqrt{1 - \frac{\phi^{2}}{4}}} \arctan y + c = \\
= \frac{1}{2} \ln \left( \left( \frac{x + \frac{\phi}{2}}{\sqrt{1 - \frac{\phi^{2}}{4}}} \right)^{2} + 1 \right) + \frac{1 - \frac{\phi}{2}}{\sqrt{1 - \frac{\phi^{2}}{4}}} \arctan \frac{x + \frac{\phi}{2}}{\sqrt{1 - \frac{\phi^{2}}{4}}} + c = \\
= \frac{1}{2} \ln \left( \frac{x^{2} + \phi x + 1}{1 - \frac{\phi^{2}}{4}} \right) + \frac{1 - \frac{\phi}{2}}{\sqrt{1 - \frac{\phi^{2}}{4}}} \arctan \frac{x + \frac{\phi}{2}}{\sqrt{1 - \frac{\phi^{2}}{4}}} + c;
\end{array} = ∫ y 2 + 1 y d y + 1 − 4 ϕ 2 1 − 2 ϕ ∫ y 2 + 1 d y = 2 1 ∫ y 2 + 1 d ( y 2 ) + 1 − 4 ϕ 2 1 − 2 ϕ arctan y = = 2 1 ln ( y 2 + 1 ) + 1 − 4 ϕ 2 1 − 2 ϕ arctan y + c = = 2 1 ln ( ( 1 − 4 ϕ 2 x + 2 ϕ ) 2 + 1 ) + 1 − 4 ϕ 2 1 − 2 ϕ arctan 1 − 4 ϕ 2 x + 2 ϕ + c = = 2 1 ln ( 1 − 4 ϕ 2 x 2 + ϕ x + 1 ) + 1 − 4 ϕ 2 1 − 2 ϕ arctan 1 − 4 ϕ 2 x + 2 ϕ + c ; c c c
∫ x + 1 x 2 − 1 ϕ x + 1 d x = ∫ x + 1 ( x − 1 2 ϕ ) 2 + 1 − 1 4 ϕ 2 d x = [ y = x − 1 2 ϕ 1 − 1 4 ϕ 2 ] = = ∫ 1 − 1 4 ϕ 2 y + 1 + 1 2 ϕ ( 1 − 1 4 ϕ 2 ) ( y 2 + 1 ) ⋅ 1 − 1 4 ϕ 2 d y = y + 1 + 1 2 ϕ 1 − 1 4 ϕ 2 y 2 + 1 d y = ∫ y y 2 + 1 d y + 1 + 1 2 ϕ 1 − 1 4 ϕ 2 ∫ d y y 2 + 1 = = 1 2 ln ( y 2 + 1 ) + 1 + 1 2 ϕ 1 − 1 4 ϕ 2 arctan y + c = = 1 2 ln ( ( x − 1 2 ϕ 1 − 1 4 ϕ 2 ) 2 + 1 ) + 1 + 1 2 ϕ 1 − 1 4 ϕ 2 arctan x − 1 2 ϕ 1 − 1 4 ϕ 2 + c = \begin{array}{l}
\int \frac{x + 1}{x^{2} - \frac{1}{\phi} x + 1} \, dx = \int \frac{x + 1}{\left( x - \frac{1}{2\phi} \right)^{2} + 1 - \frac{1}{4\phi^{2}}} \, dx = \left[ y = \frac{x - \frac{1}{2\phi}}{\sqrt{1 - \frac{1}{4\phi^{2}}}} \right] = \\
= \int \frac{ \sqrt{1 - \frac{1}{4\phi^{2}}} y + 1 + \frac{1}{2\phi} }{ \left(1 - \frac{1}{4\phi^{2}}\right) (y^{2} + 1) } \cdot \sqrt{1 - \frac{1}{4\phi^{2}}} \, dy = \\
y + \frac{ \frac{1 + \frac{1}{2\phi}}{\sqrt{1 - \frac{1}{4\phi^{2}}}} }{ y^{2} + 1 } \, dy = \int \frac{y}{y^{2} + 1} \, dy + \frac{1 + \frac{1}{2\phi}}{\sqrt{1 - \frac{1}{4\phi^{2}}}} \int \frac{dy}{y^{2} + 1} = \\
= \frac{1}{2} \ln(y^{2} + 1) + \frac{1 + \frac{1}{2\phi}}{\sqrt{1 - \frac{1}{4\phi^{2}}}} \arctan y + c = \\
= \frac{1}{2} \ln \left( \left( \frac{x - \frac{1}{2\phi}}{\sqrt{1 - \frac{1}{4\phi^{2}}}} \right)^{2} + 1 \right) + \frac{1 + \frac{1}{2\phi}}{\sqrt{1 - \frac{1}{4\phi^{2}}}} \arctan \frac{x - \frac{1}{2\phi}}{\sqrt{1 - \frac{1}{4\phi^{2}}}} + c = \\
\end{array} ∫ x 2 − ϕ 1 x + 1 x + 1 d x = ∫ ( x − 2 ϕ 1 ) 2 + 1 − 4 ϕ 2 1 x + 1 d x = [ y = 1 − 4 ϕ 2 1 x − 2 ϕ 1 ] = = ∫ ( 1 − 4 ϕ 2 1 ) ( y 2 + 1 ) 1 − 4 ϕ 2 1 y + 1 + 2 ϕ 1 ⋅ 1 − 4 ϕ 2 1 d y = y + y 2 + 1 1 − 4 ϕ 2 1 1 + 2 ϕ 1 d y = ∫ y 2 + 1 y d y + 1 − 4 ϕ 2 1 1 + 2 ϕ 1 ∫ y 2 + 1 d y = = 2 1 ln ( y 2 + 1 ) + 1 − 4 ϕ 2 1 1 + 2 ϕ 1 arctan y + c = = 2 1 ln ( ( 1 − 4 ϕ 2 1 x − 2 ϕ 1 ) 2 + 1 ) + 1 − 4 ϕ 2 1 1 + 2 ϕ 1 arctan 1 − 4 ϕ 2 1 x − 2 ϕ 1 + c = = 1 2 ln ( x 2 − x ϕ + 1 1 − 1 4 ϕ 2 ) + 1 + 1 2 ϕ 1 − 1 4 ϕ 2 arctan x − 1 2 ϕ 1 − 1 4 ϕ 2 + c ; = \frac{1}{2} \ln \left(\frac{x^2 - \frac{x}{\phi} + 1}{1 - \frac{1}{4\phi^2}}\right) + \frac{1 + \frac{1}{2\phi}}{\sqrt{1 - \frac{1}{4\phi^2}}} \arctan \frac{x - \frac{1}{2\phi}}{\sqrt{1 - \frac{1}{4\phi^2}}} + c; = 2 1 ln ( 1 − 4 ϕ 2 1 x 2 − ϕ x + 1 ) + 1 − 4 ϕ 2 1 1 + 2 ϕ 1 arctan 1 − 4 ϕ 2 1 x − 2 ϕ 1 + c ; c c c
We conclude that
∫ d x x 3 + x 8 = − 1 2 x 2 − 1 5 ln ∣ x + 1 ∣ + + 1 5 ϕ ( 1 2 ln ( x 2 + ϕ x + 1 1 − ϕ 2 4 ) + 1 − ϕ 2 1 − ϕ 2 4 arctan x + ϕ 2 1 − ϕ 2 4 + c ) − − ϕ 5 ( 1 2 ln ( x 2 − x ϕ + 1 1 − 1 4 ϕ 2 ) + 1 + 1 2 ϕ 1 − 1 4 ϕ 2 arctan x − 1 2 ϕ 1 − 1 4 ϕ 2 + c ) = = − 1 2 x 2 − 1 5 ln ∣ x + 1 ∣ + 1 5 ϕ ( 1 2 ln ( x 2 + ϕ x + 1 1 − ϕ 2 4 ) + 1 − ϕ 2 1 + ϕ 2 arctan x + ϕ 2 1 − ϕ 2 4 ) − − ϕ 5 ( 1 2 ln ( x 2 − x ϕ + 1 1 − 1 4 ϕ 2 ) + 1 + 1 2 ϕ 1 − 1 2 ϕ arctan x − 1 2 ϕ 1 − 1 4 ϕ 2 ) + c = = − 1 2 x 2 − 1 5 ln ∣ x + 1 ∣ + 1 5 ϕ ( 1 2 ln ( x 2 + ϕ x + 1 ) + 1 − ϕ 2 1 + ϕ 2 arctan x + ϕ 2 1 − ϕ 2 4 ) − − ϕ 5 ( 1 2 ln ( x 2 − x ϕ + 1 ) + 1 + 1 2 ϕ 1 − 1 2 ϕ arctan x − 1 2 ϕ 1 − 1 4 ϕ 2 ) + c , \begin{aligned}
& \int \frac{dx}{x^3 + x^8} = -\frac{1}{2x^2} - \frac{1}{5} \ln |x + 1| + \\
& \quad + \frac{1}{5\phi} \left( \frac{1}{2} \ln \left( \frac{x^2 + \phi x + 1}{1 - \frac{\phi^2}{4}} \right) + \frac{1 - \frac{\phi}{2}}{\sqrt{1 - \frac{\phi^2}{4}}} \arctan \frac{x + \frac{\phi}{2}}{\sqrt{1 - \frac{\phi^2}{4}}} + c \right) - \\
& \quad - \frac{\phi}{5} \left( \frac{1}{2} \ln \left( \frac{x^2 - \frac{x}{\phi} + 1}{1 - \frac{1}{4\phi^2}} \right) + \frac{1 + \frac{1}{2\phi}}{\sqrt{1 - \frac{1}{4\phi^2}}} \arctan \frac{x - \frac{1}{2\phi}}{\sqrt{1 - \frac{1}{4\phi^2}}} + c \right) = \\
& = -\frac{1}{2x^2} - \frac{1}{5} \ln |x + 1| + \frac{1}{5\phi} \left( \frac{1}{2} \ln \left( \frac{x^2 + \phi x + 1}{1 - \frac{\phi^2}{4}} \right) + \sqrt{ \frac{1 - \frac{\phi}{2}}{1 + \frac{\phi}{2}} } \arctan \frac{x + \frac{\phi}{2}}{\sqrt{1 - \frac{\phi^2}{4}}} \right) - \\
& \quad - \frac{\phi}{5} \left( \frac{1}{2} \ln \left( \frac{x^2 - \frac{x}{\phi} + 1}{1 - \frac{1}{4\phi^2}} \right) + \sqrt{ \frac{1 + \frac{1}{2\phi}}{1 - \frac{1}{2\phi}} } \arctan \frac{x - \frac{1}{2\phi}}{\sqrt{1 - \frac{1}{4\phi^2}}} \right) + c = \\
& = -\frac{1}{2x^2} - \frac{1}{5} \ln |x + 1| + \frac{1}{5\phi} \left( \frac{1}{2} \ln (x^2 + \phi x + 1) + \sqrt{ \frac{1 - \frac{\phi}{2}}{1 + \frac{\phi}{2}} } \arctan \frac{x + \frac{\phi}{2}}{\sqrt{1 - \frac{\phi^2}{4}}} \right) - \\
& \quad - \frac{\phi}{5} \left( \frac{1}{2} \ln \left( x^2 - \frac{x}{\phi} + 1 \right) + \sqrt{ \frac{1 + \frac{1}{2\phi}}{1 - \frac{1}{2\phi}} } \arctan \frac{x - \frac{1}{2\phi}}{\sqrt{1 - \frac{1}{4\phi^2}}} \right) + c,
\end{aligned} ∫ x 3 + x 8 d x = − 2 x 2 1 − 5 1 ln ∣ x + 1∣ + + 5 ϕ 1 ⎝ ⎛ 2 1 ln ( 1 − 4 ϕ 2 x 2 + ϕ x + 1 ) + 1 − 4 ϕ 2 1 − 2 ϕ arctan 1 − 4 ϕ 2 x + 2 ϕ + c ⎠ ⎞ − − 5 ϕ ⎝ ⎛ 2 1 ln ( 1 − 4 ϕ 2 1 x 2 − ϕ x + 1 ) + 1 − 4 ϕ 2 1 1 + 2 ϕ 1 arctan 1 − 4 ϕ 2 1 x − 2 ϕ 1 + c ⎠ ⎞ = = − 2 x 2 1 − 5 1 ln ∣ x + 1∣ + 5 ϕ 1 ⎝ ⎛ 2 1 ln ( 1 − 4 ϕ 2 x 2 + ϕ x + 1 ) + 1 + 2 ϕ 1 − 2 ϕ arctan 1 − 4 ϕ 2 x + 2 ϕ ⎠ ⎞ − − 5 ϕ ⎝ ⎛ 2 1 ln ( 1 − 4 ϕ 2 1 x 2 − ϕ x + 1 ) + 1 − 2 ϕ 1 1 + 2 ϕ 1 arctan 1 − 4 ϕ 2 1 x − 2 ϕ 1 ⎠ ⎞ + c = = − 2 x 2 1 − 5 1 ln ∣ x + 1∣ + 5 ϕ 1 ⎝ ⎛ 2 1 ln ( x 2 + ϕ x + 1 ) + 1 + 2 ϕ 1 − 2 ϕ arctan 1 − 4 ϕ 2 x + 2 ϕ ⎠ ⎞ − − 5 ϕ ⎝ ⎛ 2 1 ln ( x 2 − ϕ x + 1 ) + 1 − 2 ϕ 1 1 + 2 ϕ 1 arctan 1 − 4 ϕ 2 1 x − 2 ϕ 1 ⎠ ⎞ + c , c c c
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