What is the integral of 6/(sqrt(x+1)+\sqrt{x)} ?

The integral of 6/(sqrt(x+1)+\sqrt{x)} is 3sqrt(x)+1/((sqrt(x)+\sqrt{x+1))^3}+3sqrt(x+1)+C

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integral of 6/(sqrt(x+1)+\sqrt{x)}

\int \frac{6}{x + 1 + x} d x

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

Solution steps

\int \frac{6}{x + 1 + x} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{x + 1 + x} d x

Apply u-substitution

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

= 6 \cdot \int \frac{( u + 1 ) ( u - 1 ) ( u ^{2} + 1 )}{2 u ^{4}} d u

Take the constant out:

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

= 6 \cdot \frac{1}{2} \cdot \int \frac{( u + 1 ) ( u - 1 ) ( u ^{2} + 1 )}{u ^{4}} d u

Expand

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

= 6 \cdot \frac{1}{2} \cdot \int 1 - \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

= 6 \cdot \frac{1}{2} (\int 1 d u - \int \frac{1}{u ^{4}} d u)

\int 1 d u = u

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

= 6 \cdot \frac{1}{2} (u - (- \frac{1}{3 u ^{3}}))

Substitute back

u = x + 1 + x

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 6/(sqrt(x+1)+\sqrt{x)} ?
The integral of 6/(sqrt(x+1)+\sqrt{x)} is 3sqrt(x)+1/((sqrt(x)+\sqrt{x+1))^3}+3sqrt(x+1)+C