Q70 / 180 · KCET 2008 · Mathematics
$\int^{e^{x} \cdot x^{5}} d x$ is
- A$e^{x}\left[x^{5}+5 x^{4}+20 x^{3}+60 x^{2}+120 x+120\right]+C$
- B$e^{x}\left[x^{5}-5 x^{4}-20 x^{3}-60 x^{2}-120 x-120\right]+C$
- Correct answerC$e^{x}\left[x^{5}-5 x^{4}+20 x^{3}-60 x^{2}+120 x-120\right]+C$
- D$e^{x}\left[x^{5}+5 x^{4}+20 x^{3}-60 x^{2}-120 x+120\right]+C$
Solution
Let $I=\int e^{x} \cdot x^{5} d x$
$$
\begin{aligned}
&=e^{x} \cdot x^{5}-\int e^{x} \cdot 5 x^{4} d x \\
&=e^{x} \cdot x^{5}-5\left[e^{x} \cdot x^{4}-\int e^{x} \cdot 4 x^{3} d x\right] \\
&=e^{x} \cdot x^{5}-5 e^{x} \cdot x^{4}+20\left[e^{x} \cdot x^{3}-\int e^{x} \cdot 3 x^{2} d x\right] \\
&=e^{x} \cdot x^{5}-5 e^{x} \cdot x^{4}+20 e^{x} \cdot x^{3}-60\left[e^{x} \cdot x^{2}-\right. \\
&\left.=e^{x} \cdot x^{5}-5 e^{x} \cdot x^{4}+20 e^{x} \cdot 2 x d x\right] \\
&=e^{x}\left[x^{5}-5 x^{4}+20 e^{x} \cdot x^{2}-60 x^{2}+\right.
\end{aligned}
$$
$120 \mathrm{x}-120]+\mathrm{C}$
Hence, the correct answer is option: 'C'.