Q53 / 160 · TS EAMCET 2017 · Mathematics

If $\int e^{2 x} f^{\prime}(x) d x=g(x)$, then $\int\left(e^{2 x} f(x)+e^{2 x} f^{\prime}(x)\right] d x=$

  • Correct answer
    A
    $\frac{1}{2}\left[e^{2 x} f(x)-g(x)\right]+C$
  • B
    $\frac{1}{2}\left[e^{2 x} f(x)+g(x)\right]+C$
  • C
    $\frac{1}{2}\left[e^{2 x} f(2 x)+g(x)\right]+C$
  • D
    $\frac{1}{2}\left[e^{2 x} f^{\prime}(x)+g(x)\right]+C$

Solution

We have, $$ \begin{aligned} & \int e^{2 x} f^{\prime}(x) d x=g(x) \\ & \text { Let } I=\int\left(e^{2 x} f(x)+e^{2 x} f^{\prime}(x)\right) d x \\ & \left.=f(x) \int e^{2 x} d x-\int f^{\prime}(x) \int e^{2 x} d x\right) d x+\int e^{2 x} f^{\prime}(x) d x \\ & =\frac{f(x) e^{2 x}}{2}-\frac{1}{2} \int e^{2 x} f^{\prime}(x) d x+\int e^{2 x} f^{\prime}(x) d x \\ & =\frac{e^{2 x}}{2} f(x)-\frac{1}{2} \int e^{2 x} f^{\prime}(x) d x \\ & =\frac{1}{2}\left[e^{2 x} f(x)-\int e^{2 x} f^{\prime}(x) d x\right] \\ & =\frac{1}{2}\left[e^{2 x} f(x)-g(x)\right]+C \end{aligned} $$

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

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