What is the integral of (x^2+4)^72x ?

The integral of (x^2+4)^72x is 1/8 (x^2+4)^8+C

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integral of (x^2+4)^72x

\int (x^{2} + 4)^{7} 2 x d x

\frac{1}{8} (x^{2} + 4)^{8} + C

Solution steps

\int (x^{2} + 4)^{7} \cdot 2 x d x

Take the constant out:

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

= 2 \cdot \int (x^{2} + 4)^{7} x d x

Apply u-substitution

= 2 \cdot \int \frac{u ^{7}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int u^{7} d u

Apply the Power Rule

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{8}}{8}

Substitute back

u = x^{2} + 4

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 4 ) ^{8}}{8}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 4 ) ^{8}}{8} : \frac{1}{8} (x^{2} + 4)^{8}

= \frac{1}{8} (x^{2} + 4)^{8}

Add a constant to the solution

= \frac{1}{8} (x^{2} + 4)^{8} + C

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What is the integral of (x^2+4)^72x ?
The integral of (x^2+4)^72x is 1/8 (x^2+4)^8+C