Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals2 Marks
Question
Integrate the function $\frac{\log x}{(1+\log x)^2}$ with respect to $x$.
✓
Answer
Let $ I =\int \frac{\log x}{(1+\log x)^2} d x$
Adding and subtracting $1$ from numerator
$I=\int \frac{(\log x+1-1)}{(1+\log x)^2} d x$
$I=\int \frac{1}{(1+\log x)} d x-\int \frac{d x}{(1+\log x)^2}$
$I=\int \frac{1}{(1+\log x)} \cdot 1 d x-\int \frac{d x}{(1+\log x)^2}$
Assuming $"1\ "$ as second function integrating by parts :
$I=\frac{1}{(1+\log x)} \cdot x-\int \frac{-1}{(1+\log x)^2} \cdot \frac{1}{x} \times x d x-$
$\int \frac{d x}{(1+\log x)^2}+C$
$I=\frac{x}{(1+\log x)}+\int \frac{1}{(1+\log x)^2} d x-$
$\int \frac{1}{(1+\log x)^2} d x+C$
$I=\frac{x}{(1+\log x)}+C $
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