Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\frac{\log\text{x}}{(\text{x}+1)^2}\text{dx}$
✓
Answer
Let $\text{I}=\int\frac{\log\text{x}}{(\text{x}+1)^2}\text{dx}$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, $\int\frac{\log\text{x}}{(\text{x}+1)^2}\text{dx}=-\frac{\log\text{x}}{1+\text{x}}+\log\Big|\frac{\text{x}}{\text{x}+1}\Big|+\text{C}$
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