What is the integral of (x^6+1)e^{x^7+7x} ?

The integral of (x^6+1)e^{x^7+7x} is 1/7 e^{x^7+7x}+C

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integral of (x^6+1)e^{x^7+7x}

\int (x^{6} + 1) e^{x^{7} + 7 x} d x

\frac{1}{7} e^{x^{7} + 7 x} + C

Solution steps

\int (x^{6} + 1) e^{x^{7} + 7 x} d x

Apply u-substitution

: \int \frac{e ^{u}}{7} d u

= \int \frac{e ^{u}}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{7} e^{u}

Substitute back

u = x^{7} + 7 x

= \frac{1}{7} e^{x^{7} + 7 x}

Add a constant to the solution

= \frac{1}{7} e^{x^{7} + 7 x} + C

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What is the integral of (x^6+1)e^{x^7+7x} ?
The integral of (x^6+1)e^{x^7+7x} is 1/7 e^{x^7+7x}+C