What is the integral of 3/(sqrt(x)(1+3\sqrt{x))^2} ?

The integral of 3/(sqrt(x)(1+3\sqrt{x))^2} is 3/8 sqrt(x)+C

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integral of 3/(sqrt(x)(1+3\sqrt{x))^2}

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

\frac{3}{8} x + C

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

\frac{1}{x ( 1 + 3 x ) ^{2}} = x^{- \frac{1}{2}} \frac{1}{( 1 + 3 ) ^{2}}

= 3 \cdot \int x^{- \frac{1}{2}} \frac{1}{( 1 + 3 ) ^{2}} d x

Take the constant out:

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

= 3 \cdot \frac{1}{( 1 + 3 ) ^{2}} \cdot \int x^{- \frac{1}{2}} d x

Simplify

= 3 \cdot \frac{1}{16} \cdot \int x^{- \frac{1}{2}} d x

Apply the Power Rule

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

= 3 \cdot \frac{1}{16} \cdot 2 x^{\frac{1}{2}}

Simplify

3 \cdot \frac{1}{16} \cdot 2 x^{\frac{1}{2}} : \frac{3}{8} x

= \frac{3}{8} x

Add a constant to the solution

= \frac{3}{8} x + C

\int \frac{x}{a ^{4} - x ^{4}} d x

y = 5 x - x^{2}

x \to \infty lim (2 x + \frac{1}{x})

\int_{\frac{3}{2}}^{3} 9 - x^{2} d x

\frac{d}{d x} (cos (x^{2}))

What is the integral of 3/(sqrt(x)(1+3\sqrt{x))^2} ?
The integral of 3/(sqrt(x)(1+3\sqrt{x))^2} is 3/8 sqrt(x)+C