Question

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

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

Answer 1

The correct option is B $\frac{1}{2}\left[e^{2x}f\right(x)+g(x\left)\right]+C$

$I=\int e^{2x}.f\left(x\right)dx$ + $\int e^{2x}.f^{\prime }\left(x\right)dx$

Applying integration using By parts in the function on left hand side, we get

$\int e^{2x}.f\left(x\right)dx=\frac{e^{2x}}{2}.f\left(x\right)-\int \frac{e^{2x}}{2}.f^{\prime }\left(x\right).dx$

Adding this to $\int e^{2x}.f^{\prime }\left(x\right)dx$, we get

$I=\frac{e^{2x}}{2}.f\left(x\right)+\int \frac{e^{2x}}{2}.f^{\prime }\left(x\right).dx$