What is the integral of x/(sqrt(x^2+2x+5)) ?

The integral of x/(sqrt(x^2+2x+5)) is sqrt(x^2+2x+5)-ln(1/2 |x+1+sqrt(x^2+2x+5)|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/(sqrt(x^2+2x+5))

\int \frac{x}{x ^{2} + 2 x + 5} d x

x^{2} + 2 x + 5 - ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5) + C

Solution steps

\int \frac{x}{x ^{2} + 2 x + 5} d x

Complete the square

x^{2} + 2 x + 5 : (x + 1)^{2} + 4

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

Apply u-substitution

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

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

Expand

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

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

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

\int \frac{1}{u ^{2} + 4} d u = ln (\frac{1}{2} u + 4 + u^{2})

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

Substitute back

u = x + 1

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

Simplify

(x + 1)^{2} + 4 - ln (\frac{1}{2} x + 1 + 4 + (x + 1)^{2}) : x^{2} + 2 x + 5 - ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)

= x^{2} + 2 x + 5 - ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5)

Add a constant to the solution

= x^{2} + 2 x + 5 - ln (\frac{1}{2} x + 1 + x^{2} + 2 x + 5) + C

\int \frac{1}{4 x - 3} d x

\int_{1}^{2} (1 + 2 x)^{2} d x

ln (1 - x)

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

x = y^{2} - 15, x = 2 y

What is the integral of x/(sqrt(x^2+2x+5)) ?
The integral of x/(sqrt(x^2+2x+5)) is sqrt(x^2+2x+5)-ln(1/2 |x+1+sqrt(x^2+2x+5)|)+C