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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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 + w

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

Add a constant to the solution

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

y^{^{'}} = 6 x^{2} - 3 x^{2} y

\frac{d}{d y} (x^{- 2} y^{- 5})

x \to 0 lim (\frac{4}{sin ( 5 x )})

y^{^{'}} = - x e^{- x + y}

\int \frac{( 3 x + 5 )}{x ^{2} + 2 x + 1} d x

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