What is the integral of x\sqrt[4]{3x+1} ?

The integral of x\sqrt[4]{3x+1} is 4/9 (1/9 (3x+1)^{9/4}-1/5 (3x+1)^{5/4})+C

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integral of x\sqrt[4]{3x+1}

\int x 4 3 x + 1 d x

\frac{4}{9} (\frac{1}{9} (3 x + 1)^{\frac{9}{4}} - \frac{1}{5} (3 x + 1)^{\frac{5}{4}}) + C

Solution steps

\int x 4 3 x + 1 d x

Apply u-substitution

= \int \frac{4 u ^{4} ( u ^{4} - 1 )}{9} d u

Take the constant out:

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

= \frac{4}{9} \cdot \int u^{4} (u^{4} - 1) d u

Expand

u^{4} (u^{4} - 1) : u^{8} - u^{4}

= \frac{4}{9} \cdot \int u^{8} - u^{4} 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{4}{9} (\int u^{8} d u - \int u^{4} d u)

\int u^{8} d u = \frac{u ^{9}}{9}

\int u^{4} d u = \frac{u ^{5}}{5}

= \frac{4}{9} (\frac{u ^{9}}{9} - \frac{u ^{5}}{5})

Substitute back

u = 4 3 x + 1

= \frac{4}{9} (\frac{( 4 3 x + 1 ) ^{9}}{9} - \frac{( 4 3 x + 1 ) ^{5}}{5})

Simplify

\frac{4}{9} (\frac{( 4 3 x + 1 ) ^{9}}{9} - \frac{( 4 3 x + 1 ) ^{5}}{5}) : \frac{4}{9} (\frac{1}{9} (3 x + 1)^{\frac{9}{4}} - \frac{1}{5} (3 x + 1)^{\frac{5}{4}})

= \frac{4}{9} (\frac{1}{9} (3 x + 1)^{\frac{9}{4}} - \frac{1}{5} (3 x + 1)^{\frac{5}{4}})

Add a constant to the solution

= \frac{4}{9} (\frac{1}{9} (3 x + 1)^{\frac{9}{4}} - \frac{1}{5} (3 x + 1)^{\frac{5}{4}}) + C

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What is the integral of x\sqrt[4]{3x+1} ?
The integral of x\sqrt[4]{3x+1} is 4/9 (1/9 (3x+1)^{9/4}-1/5 (3x+1)^{5/4})+C