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

The integral of \sqrt[3]{1+ln(x)}(x^{-1}) is 3/4 (1+ln(x))^{4/3}+C

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integral of \sqrt[3]{1+ln(x)}(x^{-1})

\frac{3}{4} (1 + ln (x))^{\frac{4}{3}} + C

Solution steps

Simplify

Apply u-substitution

Apply the Power Rule

: \frac{3}{4} u^{\frac{4}{3}}

= \frac{3}{4} u^{\frac{4}{3}}

Substitute back

u = 1 + ln (x)

= \frac{3}{4} (1 + ln (x))^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{4} (1 + ln (x))^{\frac{4}{3}} + C

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What is the integral of \sqrt[3]{1+ln(x)}(x^{-1}) ?
The integral of \sqrt[3]{1+ln(x)}(x^{-1}) is 3/4 (1+ln(x))^{4/3}+C