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

The integral of (x^2-1)sqrt(2x+1) is 1/7 x^2(2x+1)^{3/2}-2/35 x(2x+1)^{3/2}-11/35 (2x+1)^{3/2}+C

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integral of (x^2-1)sqrt(2x+1)

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

\frac{1}{7} x^{2} (2 x + 1)^{\frac{3}{2}} - \frac{2}{35} x (2 x + 1)^{\frac{3}{2}} - \frac{11}{35} (2 x + 1)^{\frac{3}{2}} + C

Solution steps

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

Expand

(x^{2} - 1) 2 x + 1 : x^{2} 2 x + 1 - 2 x + 1

= \int x^{2} 2 x + 1 - 2 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 x^{2} 2 x + 1 d x - \int 2 x + 1 d x

\int x^{2} 2 x + 1 d x = \frac{1}{105} (15 x^{2} - 6 x + 2) (2 x + 1)^{\frac{3}{2}}

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

= \frac{1}{105} (15 x^{2} - 6 x + 2) (2 x + 1)^{\frac{3}{2}} - \frac{1}{3} (2 x + 1)^{\frac{3}{2}}

Simplify

\frac{1}{105} (15 x^{2} - 6 x + 2) (2 x + 1)^{\frac{3}{2}} - \frac{1}{3} (2 x + 1)^{\frac{3}{2}} : \frac{1}{7} x^{2} (2 x + 1)^{\frac{3}{2}} - \frac{2}{35} x (2 x + 1)^{\frac{3}{2}} - \frac{11}{35} (2 x + 1)^{\frac{3}{2}}

= \frac{1}{7} x^{2} (2 x + 1)^{\frac{3}{2}} - \frac{2}{35} x (2 x + 1)^{\frac{3}{2}} - \frac{11}{35} (2 x + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{7} x^{2} (2 x + 1)^{\frac{3}{2}} - \frac{2}{35} x (2 x + 1)^{\frac{3}{2}} - \frac{11}{35} (2 x + 1)^{\frac{3}{2}} + C

f (x) = x^{4} e^{x}

y = 4 x^{3} - 20 x + 25

\frac{d}{d x} (15 x^{2} - 15 x^{4})

\int \frac{1}{( x - 6 ) ^{2}} d x

\frac{\partial}{\partial x} (sin (x^{2}) + x y^{3})

What is the integral of (x^2-1)sqrt(2x+1) ?
The integral of (x^2-1)sqrt(2x+1) is 1/7 x^2(2x+1)^{3/2}-2/35 x(2x+1)^{3/2}-11/35 (2x+1)^{3/2}+C