What is the integral of t/(\sqrt[4]{1+2t)} ?

The integral of t/(\sqrt[4]{1+2t)} is 2/21 (3t-2)(2t+1)^{3/4}+C

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integral of t/(\sqrt[4]{1+2t)}

\int \frac{t}{4 1 + 2 t} d t

\frac{2}{21} (3 t - 2) (2 t + 1)^{\frac{3}{4}} + C

Solution steps

\int \frac{t}{4 1 + 2 t} d t

Apply u-substitution

= \int \frac{u - 1}{4 4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{u - 1}{4 u} d u

Expand

\frac{u - 1}{4 u} : u^{\frac{3}{4}} - \frac{1}{4 u}

= \frac{1}{4} \cdot \int u^{\frac{3}{4}} - \frac{1}{4 u} d u

Apply the Sum Rule:

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

= \frac{1}{4} (\int u^{\frac{3}{4}} d u - \int \frac{1}{4 u} d u)

\int u^{\frac{3}{4}} d u = \frac{4}{7} u^{\frac{7}{4}}

\int \frac{1}{4 u} d u = \frac{4}{3} u^{\frac{3}{4}}

= \frac{1}{4} (\frac{4}{7} u^{\frac{7}{4}} - \frac{4}{3} u^{\frac{3}{4}})

Substitute back

u = 1 + 2 t

= \frac{1}{4} (\frac{4}{7} (1 + 2 t)^{\frac{7}{4}} - \frac{4}{3} (1 + 2 t)^{\frac{3}{4}})

Simplify

\frac{1}{4} (\frac{4}{7} (1 + 2 t)^{\frac{7}{4}} - \frac{4}{3} (1 + 2 t)^{\frac{3}{4}}) : \frac{2}{21} (3 t - 2) (2 t + 1)^{\frac{3}{4}}

= \frac{2}{21} (3 t - 2) (2 t + 1)^{\frac{3}{4}}

Add a constant to the solution

= \frac{2}{21} (3 t - 2) (2 t + 1)^{\frac{3}{4}} + C

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What is the integral of t/(\sqrt[4]{1+2t)} ?
The integral of t/(\sqrt[4]{1+2t)} is 2/21 (3t-2)(2t+1)^{3/4}+C