Question

$\int \frac{xTa{n}^{-1}x}{(1+x^2{)}^{3{/}_2}}dx=$

• A. $\frac{x+Ta{n}^{-1}x}{(1+x^2{)}^3/2}+c$
• B. $\frac{x-Ta{n}^{-1}x}{\sqrt{(1+x^2)}}+c$
• C. $\frac{x}{\sqrt{1+x^2}}-Ta{n}^{-1}x+c$
• D. $\frac{x}{1+x^2}+Ta{n}^{-1}x+c$

Answer 1

The correct option is B $\frac{x-Ta{n}^{-1}x}{\sqrt{(1+x^2)}}+c$

$\int \frac{xTa{n}^{-1}x}{(1+x^2{)}^{\frac{3}{2}}}dx$
$=Ta{n}^{-1}x\int \frac{x}{(1+x^2{)}^{\frac{3}{2}}}-\int \frac{1}{(1+x^2)}[\int \frac{x}{(1+x^2{)}^{\frac{3}{2}}}dx]dx$
$=\frac{-1}{2}Ta{n}^{-1}\sqrt{1+x^2}+\int \frac{1}{(1+x^2{)}^{\frac{3}{2}}}dx$