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

The integral of x/((1+x)^{3/4)} is 4/5 (1+x)^{5/4}-4(1+x)^{1/4}+C

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integral of x/((1+x)^{3/4)}

\int \frac{x}{( 1 + x ) ^{\frac{3}{4}}} d x

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

Solution steps

\int \frac{x}{( 1 + x ) ^{\frac{3}{4}}} d x

Apply u-substitution

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

Expand

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

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

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

\int u^{\frac{1}{4}} d u = \frac{4}{5} u^{\frac{5}{4}}

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

= \frac{4}{5} u^{\frac{5}{4}} - 4 u^{\frac{1}{4}}

Substitute back

u = 1 + x

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

Add a constant to the solution

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

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