What is the integral of x/(sqrt(x^2+4)) ?

The integral of x/(sqrt(x^2+4)) is sqrt(x^2+4)+C

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integral of x/(sqrt(x^2+4))

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

x^{2} + 4 + C

Solution steps

\int \frac{x}{x ^{2} + 4} 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 = x^{2} + 4

= \frac{1}{2} \cdot 2 (x^{2} + 4)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (x^{2} + 4)^{\frac{1}{2}} : x^{2} + 4

= x^{2} + 4

Add a constant to the solution

= x^{2} + 4 + C

x \to 2 lim (- \frac{1}{( x - 2 ) ^{2}})

y = x^{2} - x + 7, at x = 7

x \to 1 lim (\frac{x ^{9} - 1}{x - 1})

\frac{d}{d x} (tan (\frac{1}{2} x))

\int sin (x + 3) d x

What is the integral of x/(sqrt(x^2+4)) ?
The integral of x/(sqrt(x^2+4)) is sqrt(x^2+4)+C