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

The integral of 1/(3+sqrt(2x-1)) is 3+sqrt(2x-1)-3ln|3+sqrt(2x-1)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(3+sqrt(2x-1))

\int \frac{1}{3 + 2 x - 1} d x

3 + 2 x - 1 - 3 ln 3 + 2 x - 1 + C

Solution steps

\int \frac{1}{3 + 2 x - 1} d x

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{3 + u} d u

Apply u-substitution

: \int \frac{2 ( v - 3 )}{v} d v

= \frac{1}{2} \cdot \int \frac{2 ( v - 3 )}{v} d v

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 \frac{v - 3}{v} d v

Expand

\frac{v - 3}{v} : 1 - \frac{3}{v}

= \frac{1}{2} \cdot 2 \cdot \int 1 - \frac{3}{v} 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

= \frac{1}{2} \cdot 2 (\int 1 d v - \int \frac{3}{v} d v)

\int 1 d v = v

\int \frac{3}{v} d v = 3 ln ∣ v ∣

= \frac{1}{2} \cdot 2 (v - 3 ln ∣ v ∣)

Substitute back

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

Simplify

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

= 3 + 2 x - 1 - 3 ln 3 + 2 x - 1

Add a constant to the solution

= 3 + 2 x - 1 - 3 ln 3 + 2 x - 1 + C

\int x^{9} ln (2 x) d x

\int sin (38 x) sin (cos (38 x)) d x

\int 7 x^{2} + 5 + \frac{9}{x ^{2} + 1} d x

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

\int \frac{1}{( 4 x ^{2} - 4 x + 5 ) ^{2}} d x

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