What is the integral of 8x(4x^2-3)^5 ?

The integral of 8x(4x^2-3)^5 is 1/6 (4x^2-3)^6+C

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integral of 8x(4x^2-3)^5

\int 8 x (4 x^{2} - 3)^{5} d x

\frac{1}{6} (4 x^{2} - 3)^{6} + C

Solution steps

\int 8 x (4 x^{2} - 3)^{5} d x

Take the constant out:

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

= 8 \cdot \int x (4 x^{2} - 3)^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{8} d u

= 8 \cdot \int \frac{u ^{5}}{8} d u

Take the constant out:

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

= 8 \cdot \frac{1}{8} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= 8 \cdot \frac{1}{8} \cdot \frac{u ^{6}}{6}

Substitute back

u = 4 x^{2} - 3

= 8 \cdot \frac{1}{8} \cdot \frac{( 4 x ^{2} - 3 ) ^{6}}{6}

Simplify

8 \cdot \frac{1}{8} \cdot \frac{( 4 x ^{2} - 3 ) ^{6}}{6} : \frac{1}{6} (4 x^{2} - 3)^{6}

= \frac{1}{6} (4 x^{2} - 3)^{6}

Add a constant to the solution

= \frac{1}{6} (4 x^{2} - 3)^{6} + C

\int \frac{1}{( x + 5 ) ^{2} ( x - 1 )} d x

\int \frac{x + 1}{x ^{2} + 4} d x

\int \frac{x}{x ^{2} + 4 x + 13} d x

x - 1

4 x

What is the integral of 8x(4x^2-3)^5 ?
The integral of 8x(4x^2-3)^5 is 1/6 (4x^2-3)^6+C