What is the integral of (t+1)^{1/3} ?

The integral of (t+1)^{1/3} is 3/4 (t+1)^{4/3}+C

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integral of (t+1)^{1/3}

\int (t + 1)^{\frac{1}{3}} d t

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

Solution steps

\int (t + 1)^{\frac{1}{3}} d t

Apply u-substitution

: \int u^{\frac{1}{3}} d u

= \int u^{\frac{1}{3}} d u

Apply the Power Rule

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

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

Substitute back

u = t + 1

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

Add a constant to the solution

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

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What is the integral of (t+1)^{1/3} ?
The integral of (t+1)^{1/3} is 3/4 (t+1)^{4/3}+C