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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int (3 1 + ln (x)) (x^{- 1}) d x

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

Solution steps

\int 3 1 + ln (x) x^{- 1} d x

Simplify

3 1 + ln (x) x^{- 1} : \frac{3 1 + ln ( x )}{x}

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

Apply u-substitution

: \int 3 u d u

= \int 3 u d u

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

\int \frac{x ^{2}}{441 - x ^{2}} d x

\int p (p + 3)^{5} d p

\int 9 sec^{6} (t) d t

\int_{0}^{\frac{π}{2}} cos^{3} (3 x) d x

(sinh (t))^{^{'}}

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