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

The integral of (sqrt(x-1))/(x+3) is 4(-arctan(1/2 sqrt(x-1))+1/2 sqrt(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 (sqrt(x-1))/(x+3)

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

4 (- arctan (\frac{1}{2} x - 1) + \frac{1}{2} x - 1) + C

Solution steps

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

Apply u-substitution

: \int \frac{2 u ^{2}}{u ^{2} + 4} d u

= \int \frac{2 u ^{2}}{u ^{2} + 4} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2}}{u ^{2} + 4} d u

\frac{u ^{2}}{u ^{2} + 4} = - \frac{4}{u ^{2} + 4} + 1

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

Apply Integral Substitution

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

Take the constant out:

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

= 2 \cdot 2 \cdot \int - \frac{1}{v ^{2} + 1} + 1 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

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

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

\int 1 d v = v

= 2 \cdot 2 (- arctan (v) + v)

Substitute back

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

Simplify

2 \cdot 2 (- arctan (\frac{x - 1}{2}) + \frac{x - 1}{2}) : 4 (- arctan (\frac{1}{2} x - 1) + \frac{1}{2} x - 1)

= 4 (- arctan (\frac{1}{2} x - 1) + \frac{1}{2} x - 1)

Add a constant to the solution

= 4 (- arctan (\frac{1}{2} x - 1) + \frac{1}{2} x - 1) + C

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

\int (5 x^{2} + 4 x + \frac{1}{x ^{3}}) d x

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

\int \frac{1}{3 5 x - 2} d x

\int \frac{6 x + 1}{( x + 2 ) ^{2}} d x

What is the integral of (sqrt(x-1))/(x+3) ?
The integral of (sqrt(x-1))/(x+3) is 4(-arctan(1/2 sqrt(x-1))+1/2 sqrt(x-1))+C