What is the integral of (-7x)/(\sqrt[4]{x+2)} ?

The integral of (-7x)/(\sqrt[4]{x+2)} is -4(x+2)^{7/4}+56/3 (x+2)^{3/4}+C

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integral of (-7x)/(\sqrt[4]{x+2)}

- 4 (x + 2)^{\frac{7}{4}} + \frac{56}{3} (x + 2)^{\frac{3}{4}} + C

Solution steps

Take the constant out:

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

Apply u-substitution

Expand

Apply the Sum Rule:

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

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

= - 7 (\frac{4}{7} u^{\frac{7}{4}} - \frac{8}{3} u^{\frac{3}{4}})

Substitute back

u = x + 2

= - 7 (\frac{4}{7} (x + 2)^{\frac{7}{4}} - \frac{8}{3} (x + 2)^{\frac{3}{4}})

Simplify

- 7 (\frac{4}{7} (x + 2)^{\frac{7}{4}} - \frac{8}{3} (x + 2)^{\frac{3}{4}}) : - 4 (x + 2)^{\frac{7}{4}} + \frac{56}{3} (x + 2)^{\frac{3}{4}}

= - 4 (x + 2)^{\frac{7}{4}} + \frac{56}{3} (x + 2)^{\frac{3}{4}}

Add a constant to the solution

= - 4 (x + 2)^{\frac{7}{4}} + \frac{56}{3} (x + 2)^{\frac{3}{4}} + C

\frac{x}{( x + 1 )}

\int (x^{2} + 4)^{\frac{1}{2}} d x

\int_{0}^{y} 0.125 d x

x \to 1 lim (x^{(\frac{1}{1 - x})})

n = 1 \sum \infty n^{2} (\frac{2}{3})^{n}

What is the integral of (-7x)/(\sqrt[4]{x+2)} ?
The integral of (-7x)/(\sqrt[4]{x+2)} is -4(x+2)^{7/4}+56/3 (x+2)^{3/4}+C