Indefinite Integrals — Maths STD 12 Science — Question

Let the first function be (log x) and second function be $\frac{1}{(\text{x}+1)^2}.$

First we find the integral of the second function, i.e, $\int\frac{1}{(\text{x}+1)^2}\text{dx}.$

Put t = (x + 1)  Then dt = dx

Therefore,

$\int\frac{1}{(\text{x}+1)^2}\text{dx}=\int\text{t}^{-2}\text{dt}$

$=-\frac{1}{\text{t}}$

$=-\frac{1}{1+\text{x}}$

Hence, using integration by parts, we get

$\int\frac{\log\text{x}}{(\text{x}+1)^2}\text{dx}=(\log\text{x})\int\frac{1}{(\text{x}+1)^2}\text{dx}-\int\Big[\Big(\frac{\text{d}(\log\text{x})}{\text{dx}}\Big)\int\frac{1}{(\text{x}+1)^2}\text{dx}\Big]\text{dx}$

$=(\log\text{x})\Big(-\frac{1}{1+\text{x}}\Big)-\int\big(\frac{1}{\text{x}}\big)\Big(-\frac{1}{1+\text{x}}\Big)\text{dx}$

$=-\frac{\log\text{x}}{1+\text{x}}+\int\Big(\frac{1}{\text{x}^2+\text{x}}\Big)\text{dx}$

$=-\frac{\log\text{x}}{1+\text{}x}+\int\frac{1}{\text{x}^2+\text{x}+\frac{1}{4}-\frac{1}{4}}\text{dx}$

$=-\frac{\log\text{x}}{1+\text{x}}+\int\frac{1}{\big(\text{x}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\text{dx}$

$=-\frac{\log\text{x}}{1+\text{x}}+\frac{1}{2\times\frac{1}{2}}\log\Bigg|\frac{\text{x}+\frac{1}{2}-\frac{1}{2}}{\text{x}+\frac{1}{2}+\frac{1}{2}}\Bigg|+\text{C}$

$=-\frac{\log\text{x}}{1+\text{x}}+\log\Big|\frac{\text{x}}{\text{x}+1}\Big|+\text{C}$

Hence, $\int\frac{\log\text{x}}{(\text{x}+1)^2}\text{dx}=-\frac{\log\text{x}}{1+\text{x}}+\log\Big|\frac{\text{x}}{\text{x}+1}\Big|+\text{C}$