Answer to Question #110639 in Calculus for MANNU KUMAR GUPTA

Question

Integrated ×^2+×+5/(×^2+4)(×+1) dx.
Integration by part

Answer 1

Problem: $\int {\frac {x^2+x+5}{(x^2+4)(x+1)}dx}$

Solution:

(1) $\int {\frac {x^2+x+5}{(x^2+4)(x+1)}dx}=\int {\frac {x^2+4+x+1}{(x^2+4)(x+1)}dx}=I_1+I_2$

(2) $I_1=\int {\frac {1}{x+1}dx}=ln(x+1)+C$

$I_2=\int {\frac {1}{x^2+4}\cdot dx}$

To evaluate $I_2$ one can use the substitution

(3) $x=2tan(y)$ then

$x^2+4=4(tan^2(y)+1)=4(\frac{sin^2(y)+cos^2(y)}{cos^2(y)})=\frac{4}{cos^2(y)};\\ dx=\frac{2dy}{cos^2(y)}$ and

(4) $I_2=\int{\frac{1}{4/cos^2(y)}\cdot \frac{2dy}{cos^2(y)}}=\frac{1}{2}\int{dy}=\frac{y}{2}+C$

Reverse substitution (3) we have $y=arctan(\frac{x}{2})$.

In notation  John Herschel we may write $y= tan^{-1}(\frac{x}{2})$.

(5) $I_2=\frac{1}{2}arctan(\frac{x}{2})+C$

Substitute (2) and (5) in (1) we get the solution to the problem

Answer:$\int {\frac {x^2+x+5}{(x^2+4)(x+1)}dx}=ln(x+1)+\frac{1}{2}arctan(\frac{x}{2})+C$

Comment:

(1) The solution method used here is not called "integration by parts", although the integral is divided into two parts for integration. The correct name for this method is 'fractional part integration" or 'Integration using Partial Fractions'. This method used for rational function integrals http://www.math.wsu.edu/faculty/genz/140/lessons/l506.pdf.

(2) If we use $y=arctan(\frac{x}{2})$ in (3) we can immediately get

$dy=arctan'_x(\frac{x}{2})dx=\frac{dx}{2}\cdot \frac{1}{1+(x/2)^2}=\frac{2dx}{4+x^2}$ and

$\frac{dx}{4+x^2}=\frac{dy}{2}$