What is the integral of x^3(7+x^4)^6 ?

The integral of x^3(7+x^4)^6 is 1/28 (7+x^4)^7+C

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integral of x^3(7+x^4)^6

\int x^{3} (7 + x^{4})^{6} d x

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

Solution steps

\int x^{3} (7 + x^{4})^{6} d x

Apply u-substitution

= \int \frac{u ^{6}}{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^{6} d u

Apply the Power Rule

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

Substitute back

u = 7 + x^{4}

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of x^3(7+x^4)^6 ?
The integral of x^3(7+x^4)^6 is 1/28 (7+x^4)^7+C