What is the integral of x^3(9+x^4)^7 ?

The integral of x^3(9+x^4)^7 is 1/32 (9+x^4)^8+C

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integral of x^3(9+x^4)^7

\int x^{3} (9 + x^{4})^{7} d x

\frac{1}{32} (9 + x^{4})^{8} + C

Solution steps

\int x^{3} (9 + x^{4})^{7} d x

Apply u-substitution

: \int \frac{u ^{7}}{4} d u

= \int \frac{u ^{7}}{4} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{8}}{8}

= \frac{1}{4} \cdot \frac{u ^{8}}{8}

Substitute back

u = 9 + x^{4}

= \frac{1}{4} \cdot \frac{( 9 + x ^{4} ) ^{8}}{8}

Simplify

\frac{1}{4} \cdot \frac{( 9 + x ^{4} ) ^{8}}{8} : \frac{1}{32} (9 + x^{4})^{8}

= \frac{1}{32} (9 + x^{4})^{8}

Add a constant to the solution

= \frac{1}{32} (9 + x^{4})^{8} + C

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What is the integral of x^3(9+x^4)^7 ?
The integral of x^3(9+x^4)^7 is 1/32 (9+x^4)^8+C