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

The integral of 2/(sqrt(x)(1+2\sqrt{x))^4} is 4/81 sqrt(x)+C

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

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

\frac{4}{81} x + C

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

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

Take the constant out:

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

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

Simplify

= 2 \cdot \frac{1}{81} \cdot \int x^{- \frac{1}{2}} d x

Apply the Power Rule

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

= 2 \cdot \frac{1}{81} \cdot 2 x^{\frac{1}{2}}

Simplify

2 \cdot \frac{1}{81} \cdot 2 x^{\frac{1}{2}} : \frac{4}{81} x

= \frac{4}{81} x

Add a constant to the solution

= \frac{4}{81} x + C

f^{^{'}} (x) = 0

\int_{- 1}^{1} e^{y} - (y^{2} - 8) d y

f (x) = \frac{1}{x}, at x = - 4

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

x (1 - 9 x^{3})

What is the integral of 2/(sqrt(x)(1+2\sqrt{x))^4} ?
The integral of 2/(sqrt(x)(1+2\sqrt{x))^4} is 4/81 sqrt(x)+C