$\int\limits_{0}^{2}\frac{x^3}{(x^2+1)^{3/2}}dx$ is equal to |
$(\sqrt{2}-1)^2$ $\frac{(\sqrt{2}-1)^2}{\sqrt{2}}$ $\frac{\sqrt{2}-1}{\sqrt{2}}$ none of these |
none of these |
Let $I=\int\limits_{0}^{2}\frac{x^3}{(x^2+1)^{3/2}}dx$ $I=\frac{1}{2}\int\limits_{1}^{5}\frac{(t-1)}{t^{3/2}}dt$, where $x^2+1=t$ $⇒I=\frac{1}{2}\int\limits_{1}^{5}(t^{-1/2}-t^{-3/2})dt$ $⇒I=\frac{1}{2}[2t^{1/2}+2t^{-1/2}]_{1}^{5}=(\sqrt{5}+\frac{1}{\sqrt{5}}-2)=\frac{6-2\sqrt{5}}{\sqrt{5}}$ |