Question

$\int \frac{a}{\left(1+x^2\right)ta{n}^{-1}x}dx=$

• A. $alog\left|ta{n}^{-1}x\right|+C$
• B. $\frac{a}{2}{\left(ta{n}^{-1}x\right)}^2+C$
• C. $alog\left(1+x^2\right)+C$
• D. None of these

Answer 1

Answer: (A)

$I=a\int \frac{dx}{\left(1+x^2\right)ta{n}^{-1}x}$

Put $ta{n}^{-1}x=t\frac{\Rightarrow 1}{1+x^2}dx=dt$

$\Rightarrow I=a\int \frac{dt}{t}=alog\left|t\right|+C=alog\left|ta{n}^{-1}x\right|+C$