What is the integral of (x-1)/(\sqrt[5]{9x^2+12x+4)} ?

The integral of (x-1)/(\sqrt[5]{9x^2+12x+4)} is 1/216 (15(3x+2)^{8/5}-200(3x+2)^{3/5})+C

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integral of (x-1)/(\sqrt[5]{9x^2+12x+4)}

\frac{1}{216} (15 (3 x + 2)^{\frac{8}{5}} - 200 (3 x + 2)^{\frac{3}{5}}) + C

Solution steps

Complete the square

9 x^{2} + 12 x + 4 : 9 (x + \frac{2}{3})^{2}

Apply u-substitution

Take the constant out:

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

Expand

\frac{3 u - 5}{u ^{\frac{2}{5}}} : 3 u^{\frac{3}{5}} - \frac{5}{u ^{\frac{2}{5}}}

Apply the Sum Rule:

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

\int 3 u^{\frac{3}{5}} d u = \frac{15}{8} u^{\frac{8}{5}}

\int \frac{5}{u ^{\frac{2}{5}}} d u = \frac{25}{3} u^{\frac{3}{5}}

Substitute back

u = x + \frac{2}{3}

Simplify

= \frac{1}{216} (15 (3 x + 2)^{\frac{8}{5}} - 200 (3 x + 2)^{\frac{3}{5}})

Add a constant to the solution

= \frac{1}{216} (15 (3 x + 2)^{\frac{8}{5}} - 200 (3 x + 2)^{\frac{3}{5}}) + C

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What is the integral of (x-1)/(\sqrt[5]{9x^2+12x+4)} ?
The integral of (x-1)/(\sqrt[5]{9x^2+12x+4)} is 1/216 (15(3x+2)^{8/5}-200(3x+2)^{3/5})+C