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

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

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

\int \frac{x - 1}{2 x + x + 1} d x

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

Solution steps

\int \frac{x - 1}{2 x + x + 1} d x

Apply u-substitution

: \int \frac{u - 2}{2 u - 1 + u} d u

= \int \frac{u - 2}{2 u - 1 + u} d u

Apply u-substitution

: \int \frac{v}{2 v + 1 + v + 2} d v

= \int \frac{v}{2 v + 1 + v + 2} d v

Multiply by the conjugate of

2 v + 1 + v + 2 : \frac{v}{2 v + 1 - v + 2}

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

Simplify

\frac{v}{\frac{v}{2 v + 1 - v + 2}} : 2 v + 1 - v + 2

= \int 2 v + 1 - v + 2 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

= \int 2 v + 1 d v - \int v + 2 d v

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

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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