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Integration by Parts

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Question

Evaluate:

$\int e^{x^{3} + x^{2} - 1} (3 x^{4} + 2 x^{3} + 2 x) d x$

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Solution

$\int e^{x^{3} + x^{2} - 1} (3 x^{4} + 2 x^{3} + 2 x) d x$
Notice that $(x^{3} + x^{2} - 1) = 3 x^{2} + 2 x$ that $3 x^{4} + 2 x^{3} = x^{2} (3 x^{2} + 2 x)$
$(3 x^{4} + 2 x^{3} + 2 x) e^{x^{3} + x^{2} - 1} = x^{2} (3 x^{2} + 2 x) e^{x^{3} + x^{2} - 1} + 2 x e^{x^{3} + x^{2} - 1}$
$= x^{2} (x^{3} + x^{2} - 1)^{1} e^{x^{3}} + x^{2} - 1 + (x^{2})^{1} e^{x^{3} + x^{2} - 1}$
$= x^{2} (e^{x^{3} + x^{2} - 1}) + (n^{2})^{1} e^{x^{3}} + x^{2} = 1$
$= (x^{2} e^{x^{3} + x^{2} - 1})^{1}$
Now, $\int (x^{2} e^{x^{3} + x^{2} - 1})^{1} d x$
$= x^{2} e^{x^{3} + x^{2} - 1} + C$

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