How do you integrate $(x^{7}) e^{x^{8}} d x$ $(x^(7))e^(x^(8)) dx$ ?

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How do you integrate $(x^{7}) e^{x^{8}} d x$ $(x^(7))e^(x^(8)) dx$ ?

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Answer 1 · 2015-04-17T16:19:55

$\int x^{7} e^{x^{8}} d x$ $int x^7e^(x^8) dx$

Use substitution to write this an an integral of $e^{u} d u$ $e^u du$

Let $u = x^{8}$ $u = x^8$ . This makes $d u = 8 x^{7} d x$ $du = 8x^7 dx$ ,

$\int x^{7} e^{x^{8}} d x = \frac{1}{8} \int e^{x^{8}} (8 x^{7}) d x = \frac{1}{8} \int e^{u} d u$ $int x^7e^(x^8) dx = 1/8 int e^(x^8)(8x^7) dx = 1/8 int e^u du$

$= \frac{1}{8} e^{u} + C = \frac{1}{8} e^{x^{8}} + C$ $=1/8 e^u +C = 1/8 e^(x^8) +C$

(Check by differentiating $\frac{1}{8} e^{x^{8}} + C$ $1/8 e^(x^8) +C$ )

Answer 2 · 2015-04-17T16:27:34

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