What is the integral of (8x^2+1)^2(16x) ?

The integral of (8x^2+1)^2(16x) is 1/3 (8x^2+1)^3+C

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integral of (8x^2+1)^2(16x)

\int (8 x^{2} + 1)^{2} (16 x) d x

\frac{1}{3} (8 x^{2} + 1)^{3} + C

Solution steps

\int (8 x^{2} + 1)^{2} \cdot 16 x d x

Take the constant out:

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

= 16 \cdot \int (8 x^{2} + 1)^{2} x d x

Apply u-substitution

= 16 \cdot \int \frac{u ^{2}}{16} d u

Take the constant out:

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

= 16 \cdot \frac{1}{16} \cdot \int u^{2} d u

Apply the Power Rule

= 16 \cdot \frac{1}{16} \cdot \frac{u ^{3}}{3}

Substitute back

u = 8 x^{2} + 1

= 16 \cdot \frac{1}{16} \cdot \frac{( 8 x ^{2} + 1 ) ^{3}}{3}

Simplify

16 \cdot \frac{1}{16} \cdot \frac{( 8 x ^{2} + 1 ) ^{3}}{3} : \frac{1}{3} (8 x^{2} + 1)^{3}

= \frac{1}{3} (8 x^{2} + 1)^{3}

Add a constant to the solution

= \frac{1}{3} (8 x^{2} + 1)^{3} + C

What is the integral of (8x^2+1)^2(16x) ?
The integral of (8x^2+1)^2(16x) is 1/3 (8x^2+1)^3+C