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

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

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

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

Solution steps

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

Factor

\frac{x}{x ^{2} + x}

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

x (x + 1) = x (x + 1)

= \int \frac{x}{x x + 1} d x

Simplify

\frac{x}{x x + 1} : \frac{x}{x + 1}

= \int \frac{x}{x + 1} d x

Apply u-substitution

: \int \frac{u - 1}{u} d u

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

Apply Integration By Parts

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

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

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

= \frac{u - 1}{u} u - ln u + u - 1

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

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

\int \frac{x ^{3}}{3 8 + x ^{4}} d x

\frac{9}{x}

\int (\frac{1}{x} - 2) d x

\int \frac{2 x}{x ^{4}} d x

\int \frac{1}{z ln ( z ) ln ( ln ( z ) )} d z

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