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

The integral of 1/((1+sqrt(x))^4) is -1/((1+sqrt(x))^2)+2/(3(1+sqrt(x))^3)+C

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integral of 1/((1+sqrt(x))^4)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2 ( u - 1 )}{u ^{4}} d u

= \int \frac{2 ( u - 1 )}{u ^{4}} 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 - 1}{u ^{4}} d u

Expand

\frac{u - 1}{u ^{4}} : \frac{1}{u ^{3}} - \frac{1}{u ^{4}}

= 2 \cdot \int \frac{1}{u ^{3}} - \frac{1}{u ^{4}} 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

= 2 (\int \frac{1}{u ^{3}} d u - \int \frac{1}{u ^{4}} d u)

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

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

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

Substitute back

u = 1 + x

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

Simplify

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

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

Add a constant to the solution

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

\int cos (x) e^{2 x} d x

3 y^{^{''}} + 48 y = 0

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

\int (x - csc (x) cot (x)) d x

f (x) = x^{2} \cdot e^{4 x}

What is the integral of 1/((1+sqrt(x))^4) ?
The integral of 1/((1+sqrt(x))^4) is -1/((1+sqrt(x))^2)+2/(3(1+sqrt(x))^3)+C