What is the integral of (8x^2-10)^6x ?

The integral of (8x^2-10)^6x is 1/112 (8x^2-10)^7+C

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integral of (8x^2-10)^6x

\int (8 x^{2} - 10)^{6} x d x

\frac{1}{112} (8 x^{2} - 10)^{7} + C

Solution steps

\int (8 x^{2} - 10)^{6} x d x

Apply u-substitution

: \int \frac{u ^{6}}{16} d u

= \int \frac{u ^{6}}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= \frac{1}{16} \cdot \frac{u ^{7}}{7}

Substitute back

u = 8 x^{2} - 10

= \frac{1}{16} \cdot \frac{( 8 x ^{2} - 10 ) ^{7}}{7}

Simplify

\frac{1}{16} \cdot \frac{( 8 x ^{2} - 10 ) ^{7}}{7} : \frac{1}{112} (8 x^{2} - 10)^{7}

= \frac{1}{112} (8 x^{2} - 10)^{7}

Add a constant to the solution

= \frac{1}{112} (8 x^{2} - 10)^{7} + C

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What is the integral of (8x^2-10)^6x ?
The integral of (8x^2-10)^6x is 1/112 (8x^2-10)^7+C