Question

Evaluate the following integrals

(1) integral of {(10x^9+40x^4+3)√(x^10+8x^5+3x+5)} DX

(2) integral of {1/[(x-5)(x^2+4)]} dx

Answer 1

1.

$\int\left(10 x^{9}+40 x^{4}+3\right) \sqrt{x^{10}+8 x^{5}+3 x+5} \mathrm{~d} x\\[4mm] \text{ Substitute } u=x^{10}+8 x^{5}+3 x+5 \longrightarrow \frac{\mathrm{d} u}{\mathrm{~d} x}=10 x^{9}+40 x^{4}+3 \longrightarrow \mathrm{d} x=\frac{1}{10 x^{9}+40 x^{4}+3} \mathrm{~d} u:\\[4mm] =\int \sqrt{u} \mathrm{~d} u\\[4mm] \text{ Apply power rule:}\\[2mm] \int u^{\mathrm{n}} \mathrm{d} u=\frac{u^{\mathbf{n}+1}}{\mathbf{n}+1} +C\text{ with }\mathbf{n}=\frac{1}{2} =\frac{2 u^{\frac{3}{2}}}{3} \text{ Substitute back } u=x^{10}+8 x^{5}+3 x+5 : \\[3mm] =\frac{2\left(x^{10}+8 x^{5}+3 x+5\right)^{\frac{3}{2}}}{3} +C\\[3mm] \text{ Hence, }\\[3mm] \begin{gathered} \int\left(10 x^{9}+40 x^{4}+3\right) \sqrt{x^{10}+8 x^{5}+3 x+5} d x \\ =\frac{2\left(x^{10}+8 x^{5}+3 x+5\right)^{\frac{3}{2}}}{3}+C \end{gathered}$

2.

$$\int \frac{1}{(x-5)\left(x^{2}+4\right)} d x\\[4mm] \text{ Perform partial fraction decomposition:}\\[2mm] =\int\left(\frac{1}{29(x-5)}-\frac{x+5}{29\left(x^{2}+4\right)}\right) d x\\[2mm] \text{ Apply linearity: } \\=\frac{1}{29} \int \frac{1}{x-5} \mathrm{~d} x-\frac{1}{29} \int \frac{x+5}{x^{2}+4} \mathrm{~d} x \\ \text{Now solving: }\\ \int \frac{1}{x-5} \mathrm{~d} x \\[4mm] \frac{\mathrm{d} u}{\mathrm{~d} x}=1 \longrightarrow \mathrm{d} x=\mathrm{d} u =\int \frac{1}{u} \mathrm{~d} u\\[2mm] \begin{gathered} =\frac{1}{29} \int \frac{1}{x-5} \mathrm{~d} x-\frac{1}{29} \int \frac{x+5}{x^{2}+4} \mathrm{~d} x \\[3mm] \text { Now solving: } \\[2mm] \int \frac{1}{x-5} \mathrm{~d} x \\ \text { Substitute } u=x-5 \longrightarrow \frac{\mathrm{d} u}{\mathrm{~d} x}=1 \longrightarrow \mathrm{d} x=\mathrm{d} u: \end{gathered}\\[2mm] =\int \frac{1}{u} \mathrm{~d} u\\[3mm] \text{ This is a standard integral:}\\[2mm] =\ln (u)+C\\[3mm] \text{ Substitute back} \,u = x-5\\ =\ln (x-5)+C\\[3mm] \text{ Now solving:}\\[2mm] \int \frac{x+5}{x^{2}+4} d x\\[3mm] \text{ Expand:}\\[2mm] =\int\left(\frac{x}{x^{2}+4}+\frac{5}{x^{2}+4}\right) d x\\ =\int \frac{x}{x^{2}+4} \mathrm{~d} x+5 \int \frac{1}{x^{2}+4} \mathrm{~d} x$$

$$\text{ Now solving:}\\[2mm] \int \frac{x}{x^{2}+4} d x\\[3mm] \text{ Substitute } u=x^{2}+4 \longrightarrow \frac{\mathrm{d} u}{\mathrm{~d} x}=2 x \longrightarrow \mathrm{d} x=\frac{1}{2 x} \mathrm{~d} u :\\[2mm] =\frac{1}{2} \int \frac{1}{u} \mathrm{~d} u\\[3mm] \text{ Now solving:}\\[2mm] \int \frac{1}{u} \mathrm{~d} u\\[2mm] \text{ Use previous result:}\\[2mm] =\ln (u)+C_1\\[2mm] \text{ Plug in solved integrals:}\\[2mm] \begin{aligned} &\frac{1}{2} \int \frac{1}{u} \mathrm{~d} u \\ &=\frac{\ln (u)}{2}+C \end{aligned} \\[2mm] \text{Substitute back } u=x^{2}+4 :\\[2mm] =\frac{\ln \left(x^{2}+4\right)}{2}+C\\[3mm] \text{ Now solving:}\\[2mm] \int \frac{1}{x^{2}+4} \mathrm{~d} x\\[2mm] \text{ Substitute } u=\frac{x}{2} \longrightarrow \frac{\mathrm{d} u}{\mathrm{~d} x}=\frac{1}{2} \longrightarrow \mathrm{d} x=2 \mathrm{~d} u :\\[2mm] =\int \frac{2}{4 u^{2}+4} d u$$

$$\text{ Simplify: }\\[2mm] =\frac{1}{2} \int \frac{1}{u^{2}+1} \mathrm{~d} u\\[2mm] \text{ Now solving:}\\[2mm] \int \frac{1}{u^{2}+1} d u\\[2mm] \text{ This is a standard integral:} =\arctan (u)+C_1\\[2mm] \text{ Plug in solved integrals:}\\[2mm] \begin{gathered} \frac{1}{2} \int \frac{1}{u^{2}+1} \mathrm{~d} u \\ =\frac{\arctan (u)}{2} +C\\ \text { Substitute back } u=\frac{x}{2} \text { : } \\ =\frac{\arctan \left(\frac{x}{2}\right)}{2}+C \end{gathered} \\[2mm] \text{ Plug in solved integrals:}\\ \begin{aligned} &\int \frac{x}{x^{2}+4} \mathrm{~d} x+5 \int \frac{1}{x^{2}+4} \mathrm{~d} x \\ &=\frac{\ln \left(x^{2}+4\right)}{2}+\frac{5 \arctan \left(\frac{x}{2}\right)}{2}+C \end{aligned}\\[2mm] \text{ Plug in solved integrals:}\\[2mm] \begin{gathered} \frac{1}{29} \int \frac{1}{x-5} \mathrm{~d} x-\frac{1}{29} \int \frac{x+5}{x^{2}+4} \mathrm{~d} x \\ =-\frac{\ln \left(x^{2}+4\right)}{58}-\frac{5 \arctan \left(\frac{x}{2}\right)}{58}+\frac{\ln (x-5)}{29}+C \end{gathered}$$

$$\text{The problem is solved. Apply the absolute value function to arguments of}\\ \text{logarithm functions in order to extend the antiderivative's domain:}\\[2mm] \begin{aligned} & \int \frac{1}{(x-5)\left(x^{2}+4\right)} d x \\ =-& \frac{\ln \left(x^{2}+4\right)}{58}-\frac{5 \arctan \left(\frac{x}{2}\right)}{58}+\frac{\ln (|x-5|)}{29}+C \\ =&-\frac{\ln \left(x^{2}+4\right)+5 \arctan \left(\frac{x}{2}\right)-2 \ln (|x-5|)}{58}+C \end{aligned}$$