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

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

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

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

1 + x^{2} + C

Solution steps

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

Apply u-substitution

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

= \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

: 2 u^{\frac{1}{2}}

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

Substitute back

u = 1 + x^{2}

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

Simplify

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

= 1 + x^{2}

Add a constant to the solution

= 1 + x^{2} + C

\frac{d}{d x} (3 x (sin (x) + cos (x)))

x \to 5 lim (\frac{3 x + 3}{x - 5})

x^{x - 1}

\frac{\partial}{\partial x} (x ln (z^{6} + x^{4}))

y = \frac{6 - x e ^{x}}{x + e ^{x}}

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