# Evaluate the following integral:

Question:

Evaluate the following integral:

$\int \frac{x^{2}+1}{\left(x^{2}+4\right)\left(x^{2}+25\right)} d x$

Solution:

Denominator is factorized, so let separate the fraction through partial fraction, hence let

$\frac{\mathrm{x}^{2}+1}{\left(\mathrm{x}^{2}+4\right)\left(\mathrm{x}^{2}+25\right)}=\frac{\mathrm{Ax}+\mathrm{B}}{\left(\mathrm{x}^{2}+4\right)}+\frac{\mathrm{Cx}+\mathrm{D}}{\mathrm{x}^{2}+25} \ldots \ldots$ (i)

$\Rightarrow \frac{\mathrm{x}^{2}+1}{\left(\mathrm{x}^{2}+4\right)\left(\mathrm{x}^{2}+25\right)}=\frac{(\mathrm{Ax}+\mathrm{B})\left(\mathrm{x}^{2}+25\right)+(\mathrm{Cx}+\mathrm{D})\left(\mathrm{x}^{2}+4\right)}{\left(\mathrm{x}^{2}+4\right)\left(\mathrm{x}^{2}+25\right)}$

$\Rightarrow \mathrm{x}^{2}+1=(\mathrm{Ax}+\mathrm{B})\left(\mathrm{x}^{2}+25\right)+(\mathrm{Cx}+\mathrm{D})\left(\mathrm{x}^{2}+4\right)$

$\Rightarrow \mathrm{x}^{2}+1=\mathrm{Ax}^{3}+25 \mathrm{Ax}+\mathrm{Bx}^{2}+25 \mathrm{~B}+\mathrm{Cx}^{3}+4 \mathrm{Cx}+\mathrm{Dx}^{2}+4 \mathrm{D}$

\begin{aligned} \Rightarrow \mathrm{x}^{2}+1=&(\mathrm{A}+\mathrm{C}) \mathrm{x}^{3}+(\mathrm{B}+\mathrm{D}) \mathrm{x}^{2}+(25 \mathrm{~A}+4 \mathrm{C}) \mathrm{x}+(25 \mathrm{~B}\\ &+4 \mathrm{D}) \ldots \ldots \text { (ii) } \end{aligned}

By equating similar terms, we get

$A+C=0 \Rightarrow A=-C \ldots \ldots \ldots \ldots$ (iii)

$B+D=1 \Rightarrow B=1-D \ldots \ldots \ldots \ldots$ (iv)

$25 A+4 C=0$

$\Rightarrow 25(-C)+4 C=0$ (from equation(iii))

$\Rightarrow C=0 \ldots \ldots \ldots .(\mathrm{v})$

$25 B+4 D=1 \Rightarrow 25(1-D)+4 D=1 \Rightarrow 21 D=24 \Rightarrow D=\frac{24}{21}=\frac{8}{7}$

So equation(iv) becomes $\mathrm{B}=1-\frac{8}{7}=-\frac{1}{7}$

So equation (iii) becomes, $A=0$

We put the values of $A, B, C$, and $D$ values back into our partial fractions in equation (i) and replace this as the integrand. We get

$\int \frac{x^{2}+1}{\left(x^{2}+4\right)\left(x^{2}+25\right)} d x$

$\Rightarrow \int\left[\frac{A x+B}{\left(x^{2}+4\right)}+\frac{C x+D}{x^{2}+25}\right] d x$

$\Rightarrow \int\left[\frac{(0) x-\frac{1}{7}}{\left(x^{2}+4\right)}+\frac{(0) x+\frac{8}{7}}{x^{2}+25}\right] d x$

Split up the integral,

$\Rightarrow-\frac{1}{7} \int \frac{1}{\left(x^{2}+4\right)} d x+\frac{8}{7} \int \frac{1}{\left(x^{2}+25\right)} d x$

Let substitute

$u=\frac{x}{2} \Rightarrow d u=\frac{1}{2} d x \Rightarrow d x=2 d u$ in first partthe

$\mathrm{v}=\frac{\mathrm{x}}{5} \Rightarrow \mathrm{dv}=\frac{1}{5} \mathrm{dx} \Rightarrow \mathrm{dx}=5 \mathrm{dv}$ in second parthe $\mathrm{t}$

so the above equation becomes,

$\Rightarrow \frac{8}{7} \int \frac{5}{\left((5 v)^{2}+25\right)} d v-\frac{1}{7} \int \frac{2}{\left((2 u)^{2}+4\right)} d u$

$\Rightarrow \frac{8}{7} \int \frac{5}{\left(25 v^{2}+25\right)} d v-\frac{1}{7} \int \frac{2}{\left(4 u^{2}+4\right)} d u$

$\Rightarrow \frac{8}{35} \int \frac{1}{v^{2}+1} d v-\frac{1}{14} \int \frac{1}{u^{2}+1} d u$

On integrating we get

$\Rightarrow \frac{8}{35} \tan ^{-1} \mathrm{v}-\frac{1}{14} \tan ^{-1} \mathrm{u}+\mathrm{C}$

(the standard integral of $\frac{1}{x^{2}+1}=\tan ^{-1} x$ )

Substituting back, we get

$\Rightarrow \frac{8}{35} \tan ^{-1}\left(\frac{x}{5}\right)-\frac{1}{14} \tan ^{-1}\left(\frac{x}{2}\right)+C$

Note: the absolute value signs account for the domain of the natural log function $(x>0)$.

Hence,

$\int \frac{\mathrm{x}^{2}+1}{\left(\mathrm{x}^{2}+4\right)\left(\mathrm{x}^{2}+25\right)} \mathrm{dx}=\frac{8}{35} \tan ^{-1}\left(\frac{\mathrm{x}}{5}\right)-\frac{1}{14} \tan ^{-1}\left(\frac{\mathrm{x}}{2}\right)+\mathrm{C}$