Evaluate the following integral:


Evaluate the following integral:

$\int \frac{1}{x\left\{6(\log x)^{2}+7 \log x+2\right\}} d x$


Let substitute $u=\log x \Rightarrow d u=\frac{1}{x} d x$, so the given equation becomes

$\int \frac{1}{x\left\{6(\log x)^{2}+7 \log x+2\right\}} d x=\int \frac{1}{\left\{6 u^{2}+7 u+2\right\}} d u \ldots$ (i)

Factorizing the denominator, we get

$\int \frac{1}{(2 u+1)(3 u+2)} d u$

The denominator is factorized, so let separate the fraction through partial fraction, hence let

$\frac{1}{(2 u+1)(3 u+2)}=\frac{A}{2 u+1}+\frac{B}{(3 u+2)} \ldots \ldots$ (ii)

$\Rightarrow \frac{1}{(2 \mathrm{u}+1)(3 \mathrm{u}+2)}=\frac{\mathrm{A}(3 \mathrm{u}+2)+\mathrm{B}(2 \mathrm{u}+1)}{(2 \mathrm{u}+1)(3 \mathrm{u}+2)}$

$\Rightarrow 1=A(3 u+2)+B(2 u+1) \ldots \ldots$ (ii)

We need to solve for $A$ and $B$. One way to do this is to pick values for $x$ which will cancel each variable.

Put $u=-\frac{2}{3}$ in the above equation, we get

$\Rightarrow 1=A\left(3\left(-\frac{2}{3}\right)+2\right)+B\left(2\left(-\frac{2}{3}\right)+1\right)$

$\Rightarrow 1=-\frac{1}{3} B$

$\Rightarrow B=-3$

Now put $u=-\frac{1}{2}$ in equation (ii), we get

$\Rightarrow 1=A\left(3\left(-\frac{1}{2}\right)+2\right)+B\left(2\left(-\frac{1}{2}\right)+1\right)$

$\Rightarrow 1=\frac{1}{2} \mathrm{~A}$

$\Rightarrow \mathrm{A}=2$

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

$\int\left[\frac{1}{(2 u+1)(3 u+2)}\right] d u$

$\Rightarrow \int\left[\frac{\mathrm{A}}{2 \mathrm{u}+1}+\frac{\mathrm{B}}{(3 \mathrm{u}+2)}\right] \mathrm{du}$

$\Rightarrow \int\left[\frac{2}{2 \mathrm{u}+1}+\frac{-3}{(3 \mathrm{u}+2)}\right] \mathrm{du}$

Split up the integral,

$\Rightarrow 2 \int \frac{1}{2 u+1} d u-3 \int\left[\frac{1}{3 u+2}\right] d u$

Let substitute

$z=2 u+1 \Rightarrow d z=2 d u$ and $y=3 u+2 \Rightarrow d y=3 d u$ so the above equation becomes,

$\Rightarrow \int \frac{1}{\mathrm{z}} \mathrm{dz}-\int\left[\frac{1}{\mathrm{y}}\right] \mathrm{dy}$

On integrating we get

$\Rightarrow \log |z|-\log |y|+C$

Substituting back the value of $z$, we get

$\Rightarrow \log |2 u+1|-\log |3 u+2|+C$

Now substitute back the value of $u$, we get

$\Rightarrow \log |2(\log x)+1|-\log |3(\log x)+2|+C$

Applying the rules of logarithm we get

$\Rightarrow \log \left|\frac{2(\log x)+1}{3(\log x)+2}\right|+c$

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


$\int \frac{1}{x\left\{6(\log x)^{2}+7 \log x+2\right\}} d x=\log \left|\frac{2(\log x)+1}{3(\log x)+2}\right|+C+C$

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