CUET Preparation Today
Find $\int x e^{(1+x^2)} dx$. |
$e^{(1+x^2)} + C$ $\frac{1}{2}e^{(1+x^2)} + C$ $2e^{(1+x^2)} + C$ $x^2e^{(1+x^2)} + C$ |
$\frac{1}{2}e^{(1+x^2)} + C$ |
The correct answer is Option (2) → $\frac{1}{2}e^{(1+x^2)} + C$ Let $I=\int x e^{(1+x^2)} dx$ Put $t = 1 + x^2$, then $dt = 2x dx ⇒\frac{dt}{2} = x dx$. $∴I = \int e^t \frac{dt}{2} = \frac{1}{2} \int e^t dt$ $= \frac{1}{2} e^t + C = \frac{1}{2} e^{(1+x^2)} + C$ |
