What is the integral of 36(4x+4)^2-2sqrt(3x-1) ?

The integral of 36(4x+4)^2-2sqrt(3x-1) is 36(16x^3-(32x^3)/3+16x^2+16x)-4/9 (3x-1)^{3/2}+C

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integral of 36(4x+4)^2-2sqrt(3x-1)

\int 36 (4 x + 4)^{2} - 2 3 x - 1 d x

36 (16 x^{3} - \frac{32 x ^{3}}{3} + 16 x^{2} + 16 x) - \frac{4}{9} (3 x - 1)^{\frac{3}{2}} + C

Solution steps

\int 36 (4 x + 4)^{2} - 2 3 x - 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 36 (4 x + 4)^{2} d x - \int 2 3 x - 1 d x

\int 36 (4 x + 4)^{2} d x = 36 (16 x^{3} - \frac{32 x ^{3}}{3} + 16 x^{2} + 16 x)

\int 2 3 x - 1 d x = \frac{4}{9} (3 x - 1)^{\frac{3}{2}}

= 36 (16 x^{3} - \frac{32 x ^{3}}{3} + 16 x^{2} + 16 x) - \frac{4}{9} (3 x - 1)^{\frac{3}{2}}

Add a constant to the solution

= 36 (16 x^{3} - \frac{32 x ^{3}}{3} + 16 x^{2} + 16 x) - \frac{4}{9} (3 x - 1)^{\frac{3}{2}} + C

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What is the integral of 36(4x+4)^2-2sqrt(3x-1) ?
The integral of 36(4x+4)^2-2sqrt(3x-1) is 36(16x^3-(32x^3)/3+16x^2+16x)-4/9 (3x-1)^{3/2}+C