What is the integral of 1/((6+sqrt(x))^3) ?

The integral of 1/((6+sqrt(x))^3) is 2(-1/(6+sqrt(x))+3/((6+sqrt(x))^2))+C

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integral of 1/((6+sqrt(x))^3)

\int \frac{1}{( 6 + x ) ^{3}} d x

2 (- \frac{1}{6 + x} + \frac{3}{( 6 + x ) ^{2}}) + C

Solution steps

\int \frac{1}{( 6 + x ) ^{3}} d x

Apply u-substitution

: \int \frac{2 u}{( 6 + u ) ^{3}} d u

= \int \frac{2 u}{( 6 + u ) ^{3}} d u

Take the constant out:

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

= 2 \cdot \int \frac{u}{( 6 + u ) ^{3}} d u

Apply u-substitution

: \int \frac{v - 6}{v ^{3}} d v

= 2 \cdot \int \frac{v - 6}{v ^{3}} d v

Expand

\frac{v - 6}{v ^{3}} : \frac{1}{v ^{2}} - \frac{6}{v ^{3}}

= 2 \cdot \int \frac{1}{v ^{2}} - \frac{6}{v ^{3}} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{1}{v ^{2}} d v - \int \frac{6}{v ^{3}} d v)

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

\int \frac{6}{v ^{3}} d v = - \frac{3}{v ^{2}}

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

Substitute back

= 2 (- \frac{1}{6 + x} - (- \frac{3}{( 6 + x ) ^{2}}))

Simplify

= 2 (- \frac{1}{6 + x} + \frac{3}{( 6 + x ) ^{2}})

Add a constant to the solution

= 2 (- \frac{1}{6 + x} + \frac{3}{( 6 + x ) ^{2}}) + C

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

f (x) = \frac{sin ( x )}{2 x}

\int_{- π}^{π} sin (x) sin (n x) d x

\frac{1}{3 + 5 x}

What is the integral of 1/((6+sqrt(x))^3) ?
The integral of 1/((6+sqrt(x))^3) is 2(-1/(6+sqrt(x))+3/((6+sqrt(x))^2))+C