Q182 / 270 · NDA 2010 (Phase 2) · Mathematics

What is $\int \sqrt{x} e^{\sqrt{x}} d x$ equal to? where $c$ is a constant of integration.

  • Correct answer
    A
    $2 e^{\sqrt{x}}(x-2 \sqrt{x}+2)+c$
  • B
    $2 e^{\sqrt{x}}(x+2 \sqrt{x}+2)+c$
  • C
    $2 e^{\sqrt{x}}(x+2 \sqrt{x}-2)+c$
  • D
    $2 e^{\sqrt{x}}(x-2 \sqrt{x}-2)+c$

Solution

$\int \sqrt{x} e^{\sqrt{x}} d x$ Put $\sqrt{x}=u$ $\begin{aligned} & \frac{1}{2 \sqrt{x}}=d u \\ & \therefore d x=2 \sqrt{x} d u \\ &= \int u^2 e^u d u \times 2 \\ &= 2\left[u^2 \int e^u d u-\int 2 u e^u d u\right] \\ &= 2\left[u^2 \cdot e^u-2\left\{e^u \cdot u-e^u\right\}\right] \\ &= 2 u^2 e^u-4 u e^u+4 e^u \\ &= 2 x e^{\sqrt{x}}-4 \sqrt{x} e^{\sqrt{x}}+4 e^{\sqrt{x}} \\ &= 2 e^{\sqrt{x}}(x-2 \sqrt{x}+2)\end{aligned}$

Hence, the correct answer is option: 'A'.

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