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

The integral of 1/(sqrt(x)(\sqrt{x)+1)^4} is 2sqrt(x)+C

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

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

2 x + C

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{x ( x + 1 ) ^{4}} = x^{- \frac{1}{2}} \frac{1}{( + 1 ) ^{4}}

= \int x^{- \frac{1}{2}} \frac{1}{1 ^{4}} d x

Simplify

x^{- \frac{1}{2}} \frac{1}{1 ^{4}} : \frac{1}{x}

= \int \frac{1}{x} d x

Apply the Power Rule

: 2 x^{\frac{1}{2}}

= 2 x^{\frac{1}{2}}

Apply radical rule:

= 2 x

Add a constant to the solution

= 2 x + C

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What is the integral of 1/(sqrt(x)(\sqrt{x)+1)^4} ?
The integral of 1/(sqrt(x)(\sqrt{x)+1)^4} is 2sqrt(x)+C