What is the integral of (4x+4)^2 ?

The integral of (4x+4)^2 is 16x^3-(32x^3)/3+16x^2+16x+C

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integral of (4x+4)^2

\int (4 x + 4)^{2} d x

16 x^{3} - \frac{32 x ^{3}}{3} + 16 x^{2} + 16 x + C

Solution steps

\int (4 x + 4)^{2} d x

Apply Integration By Parts

: (4 x + 4)^{2} x - \int 8 (4 x + 4) x d x

= (4 x + 4)^{2} x - \int 8 (4 x + 4) x d x

\int 8 (4 x + 4) x d x = 8 (\frac{4 x ^{3}}{3} + 2 x^{2})

= (4 x + 4)^{2} x - 8 (\frac{4 x ^{3}}{3} + 2 x^{2})

Simplify

(4 x + 4)^{2} x - 8 (\frac{4 x ^{3}}{3} + 2 x^{2}) : 16 x^{3} - \frac{32 x ^{3}}{3} + 16 x^{2} + 16 x

= 16 x^{3} - \frac{32 x ^{3}}{3} + 16 x^{2} + 16 x

Add a constant to the solution

= 16 x^{3} - \frac{32 x ^{3}}{3} + 16 x^{2} + 16 x + C

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What is the integral of (4x+4)^2 ?
The integral of (4x+4)^2 is 16x^3-(32x^3)/3+16x^2+16x+C