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

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

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integral of 1/(sqrt(x)(\sqrt[4]{x)+1)^5}

Solution steps

Apply u-substitution

: \int \frac{4 ( u - 1 )}{u ^{5}} d u

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

Take the constant out:

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

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

Expand

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

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

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

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

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

= 4 (- \frac{1}{3 u ^{3}} - (- \frac{1}{4 u ^{4}}))

Substitute back

Simplify

Add a constant to the solution

\int \frac{4 y}{x ^{3}} d x

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

\int 9 x (4 + 3 x)^{5} d x

\int \frac{3 y ^{2}}{x - 5} d x

\int 0.1682 x^{2} + 0.0987 x + 0.4774 d x

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