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

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

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

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

x^{2} + 1 + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{2} + 1

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

Simplify

\frac{1}{2} \cdot 2 (x^{2} + 1)^{\frac{1}{2}} : x^{2} + 1

= x^{2} + 1

Add a constant to the solution

= x^{2} + 1 + C

\frac{\partial}{\partial x} (\frac{( x + y )}{2})

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

\int_{7}^{14} 196 - x^{2} d x

s im pl i f y \frac{2 t}{2 + t}

\int \frac{x ^{3}}{x ^{4} + 9} d x

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