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

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

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

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

2 ln x + 1 + x + C

Solution steps

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

Factor

\frac{1}{x ^{2} + x}

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

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

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

= 2 ln u^{2} + 1 + u

Substitute back

u = x

= 2 ln (x)^{2} + 1 + x

(x)^{2} = x

= 2 ln x + 1 + x

Add a constant to the solution

= 2 ln x + 1 + x + C

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