What is the integral of x\sqrt[4]{3x+5} ?

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

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integral of x\sqrt[4]{3x+5}

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

Solution steps

Apply u-substitution

: \int \frac{4 u ^{4} ( u ^{4} - 5 )}{9} d u

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

Expand

u^{4} (u^{4} - 5) : u^{8} - 5 u^{4}

= \frac{4}{9} \cdot \int u^{8} - 5 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 5 u^{4} d u)

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

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

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

Substitute back

Simplify

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

Add a constant to the solution

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

\int x (x + 3)^{2} d x

(3 x^{2} + 3 x - 1)^{4}

10 \frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 6 y = 0

n = 2 \sum \infty \frac{2}{n ^{2} + 4 n + 3}

\frac{1}{∣ 1 - x ∣}

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