What is the integral of xe^{x^8} ?

The integral of xe^{x^8} is 1/2 (-(x^2\Gamma(1/4 ,-(x^2)^4))/(4\sqrt[4]{-(x^2)^4)})+C

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integral of xe^{x^8}

\int x e^{x^{8}} d x

\frac{1}{2} - \frac{x ^{2} Γ ( \frac{1}{4} , - ( x ^{2} ) ^{4} )}{4 4 - ( x ^{2} ) ^{4}} + C

Solution steps

\int x e^{x^{8}} d x

Apply u-substitution

: \int \frac{e ^{u^{4}}}{2} d u

= \int \frac{e ^{u^{4}}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int e^{u^{4}} d u

Use the nonelementary integral:

\int e^{x^{a}} d x = - \frac{x Γ ( \frac{1}{a} , - x ^{a} )}{a a - x ^{a}}

= \frac{1}{2} (- \frac{u Γ ( \frac{1}{4} , - u ^{4} )}{4 4 - u ^{4}})

Substitute back

u = x^{2}

= \frac{1}{2} - \frac{x ^{2} Γ ( \frac{1}{4} , - ( x ^{2} ) ^{4} )}{4 4 - ( x ^{2} ) ^{4}}

Add a constant to the solution

= \frac{1}{2} - \frac{x ^{2} Γ ( \frac{1}{4} , - ( x ^{2} ) ^{4} )}{4 4 - ( x ^{2} ) ^{4}} + C

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What is the integral of xe^{x^8} ?
The integral of xe^{x^8} is 1/2 (-(x^2\Gamma(1/4 ,-(x^2)^4))/(4\sqrt[4]{-(x^2)^4)})+C