What is the integral of (x^2)/(\sqrt[3]{x-1)} ?

The integral of (x^2)/(\sqrt[3]{x-1)} is x^2(x-1)^{2/3}-1/40 (25x^2-18x-27)(x-1)^{2/3}+C

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integral of (x^2)/(\sqrt[3]{x-1)}

\int \frac{x ^{2}}{3 x - 1} d x

x^{2} (x - 1)^{\frac{2}{3}} - \frac{1}{40} (25 x^{2} - 18 x - 27) (x - 1)^{\frac{2}{3}} + C

Solution steps

\int \frac{x ^{2}}{3 x - 1} d x

Apply u-substitution

: \int \frac{( u + 1 ) ^{2}}{3 u} d u

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

Apply Integration By Parts

: u^{\frac{2}{3}} (u + 1)^{2} - \int \frac{6 3 u u ^{\frac{5}{3}} + 6 3 u u ^{\frac{2}{3}} - u ^{2} - 2 u - 1}{3 u ^{\frac{1}{3}}} d u

= u^{\frac{2}{3}} (u + 1)^{2} - \int \frac{6 3 u u ^{\frac{5}{3}} + 6 3 u u ^{\frac{2}{3}} - u ^{2} - 2 u - 1}{3 u ^{\frac{1}{3}}} d u

\int \frac{6 3 u u ^{\frac{5}{3}} + 6 3 u u ^{\frac{2}{3}} - u ^{2} - 2 u - 1}{3 u ^{\frac{1}{3}}} d u = \frac{1}{40} u^{\frac{2}{3}} (25 u^{2} + 32 u - 20)

= u^{\frac{2}{3}} (u + 1)^{2} - \frac{1}{40} u^{\frac{2}{3}} (25 u^{2} + 32 u - 20)

Substitute back

u = x - 1

= (x - 1)^{\frac{2}{3}} (x - 1 + 1)^{2} - \frac{1}{40} (x - 1)^{\frac{2}{3}} (25 (x - 1)^{2} + 32 (x - 1) - 20)

Simplify

(x - 1)^{\frac{2}{3}} (x - 1 + 1)^{2} - \frac{1}{40} (x - 1)^{\frac{2}{3}} (25 (x - 1)^{2} + 32 (x - 1) - 20) : x^{2} (x - 1)^{\frac{2}{3}} - \frac{1}{40} (25 x^{2} - 18 x - 27) (x - 1)^{\frac{2}{3}}

= x^{2} (x - 1)^{\frac{2}{3}} - \frac{1}{40} (25 x^{2} - 18 x - 27) (x - 1)^{\frac{2}{3}}

Add a constant to the solution

= x^{2} (x - 1)^{\frac{2}{3}} - \frac{1}{40} (25 x^{2} - 18 x - 27) (x - 1)^{\frac{2}{3}} + C

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What is the integral of (x^2)/(\sqrt[3]{x-1)} ?
The integral of (x^2)/(\sqrt[3]{x-1)} is x^2(x-1)^{2/3}-1/40 (25x^2-18x-27)(x-1)^{2/3}+C