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Integration (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Integration (p-1)3 Marks
Question
Evalute : $\int x^2 e^{3 x} d x$
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Answer

$
\begin{aligned}
& \text { Let } I=\int x^2 e^{3 x} d x \\
& =x^2 \int e^{3 x} d x-\int\left[\frac{d}{d x}\left(x^2\right) \int e^{3 x} d x\right] d x \\
& =x^2 \cdot\left(\frac{e^{3 x}}{3}\right)-\int 2 x \cdot \frac{e^{3 x}}{3} d x \\
& =\frac{x^2}{3} e^{3 x}-\frac{2}{3} \int x \cdot e^{3 x} d x \\
& =\frac{x^2}{3} e^{3 x}-\frac{2}{3}\left[x \int e^{3 x} d x-\int\left(\frac{d}{d x}(x) \int e^{3 x} d x\right) d x\right] \\
& =\frac{x^2 \cdot e^{3 x}}{3}-\frac{2}{3}\left[x \cdot \frac{e^{3 x}}{3}-\int 1 \cdot \frac{e^{3 x}}{3} d x\right] \\
& =\frac{x^2 \cdot e^{3 x}}{3}-\frac{2}{3}\left[\frac{1}{3} x e^{3 x}-\frac{1}{3} \int e^{3 x} d x\right] \\
& =\frac{x^2 \cdot e^{3 x}}{3}-\frac{2}{3}\left[\frac{1}{3} x e^{3 x}-\frac{1}{3} \cdot \frac{e^{3 x}}{3}\right]+c\\
& \therefore I=\frac{1}{3} x^2 \cdot e^{3 x}-\frac{2}{9} x e^{3 x}+\frac{2}{27} e^{3 x}+c \\
\end{aligned}
$

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