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

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

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

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

2 ln x^{2} + 1 + x + C

Solution steps

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

Take the constant out:

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

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

Use the common integral:

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

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

Add a constant to the solution

= 2 ln x^{2} + 1 + x + C

x^{2} y^{^{'}} = x y + 3 y^{2}

\int 2 sin (t) cos (t) d t

\frac{d s}{d s}

x \to 0 lim (cos^{\frac{2}{x}} (x))

\frac{d y}{d x} = sin (9 x)

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