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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x+1)/(sqrt(2x-2))

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

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

Solution steps

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

\frac{x + 1}{2 x - 2} = \frac{1}{2} \frac{x + 1}{x - 1}

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int u^{2} + 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

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

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

\int 2 d u = 2 u

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{7 ^{x}}{1 + 7 ^{2 x}} d x

\int \frac{11}{x ^{10}} d x

\int (x^{2} + y^{2}) d x

\int \frac{1}{9 + 3 x ^{2}} d x

\int \frac{( x ^{3} + 1 )}{x ^{3} - 8} d x

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