What is the integral of (e^{7x}+1)^3e^{7x} ?

The integral of (e^{7x}+1)^3e^{7x} is 1/28 (e^{7x}+1)^4+C

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integral of (e^{7x}+1)^3e^{7x}

\int (e^{7 x} + 1)^{3} e^{7 x} d x

\frac{1}{28} (e^{7 x} + 1)^{4} + C

Solution steps

\int (e^{7 x} + 1)^{3} e^{7 x} d x

Apply u-substitution

: \int \frac{u ^{3}}{7} d u

= \int \frac{u ^{3}}{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 u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{1}{7} \cdot \frac{u ^{4}}{4}

Substitute back

u = e^{7 x} + 1

= \frac{1}{7} \cdot \frac{( e ^{7 x} + 1 ) ^{4}}{4}

Simplify

\frac{1}{7} \cdot \frac{( e ^{7 x} + 1 ) ^{4}}{4} : \frac{1}{28} (e^{7 x} + 1)^{4}

= \frac{1}{28} (e^{7 x} + 1)^{4}

Add a constant to the solution

= \frac{1}{28} (e^{7 x} + 1)^{4} + C

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What is the integral of (e^{7x}+1)^3e^{7x} ?
The integral of (e^{7x}+1)^3e^{7x} is 1/28 (e^{7x}+1)^4+C