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 u + 1 - u + 2} d u

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

Multiply by the conjugate of

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

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

Simplify

\frac{u}{\frac{u}{2 u + 1 + u + 2}} : 2 u + 1 + u + 2

= \int 2 u + 1 + u + 2 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

= \int 2 u + 1 d u + \int u + 2 d u

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

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

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

Substitute back

u = x - 1

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

Simplify

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

\frac{d}{d x} (2 x e^{- 2 x})

n = 5 \sum \infty \frac{5 ^{n}}{6 ^{n}}

\frac{d}{d x} (\frac{5}{4 x ^{3} + 4})

\int 28 (7 x - 2)^{- 5} d x

\frac{\partial}{\partial x} (\frac{( 2 x - 2 y )}{4 x + 2 y})

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