What is the integral of x/(sqrt(x^2+9)) ?

The integral of x/(sqrt(x^2+9)) is sqrt(x^2+9)+C

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integral of x/(sqrt(x^2+9))

\int \frac{x}{x ^{2} + 9} d x

x^{2} + 9 + C

Solution steps

\int \frac{x}{x ^{2} + 9} 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} + 9

= \frac{1}{2} \cdot 2 (x^{2} + 9)^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot 2 (x^{2} + 9)^{\frac{1}{2}} : x^{2} + 9

= x^{2} + 9

Add a constant to the solution

= x^{2} + 9 + C

6 t^{3} + 5 cos (2 t)

\frac{d}{d x} (x^{2} 8 x - 5)

n = 0 \sum \infty 4 (\frac{1}{10})^{n}

t^{2} y^{^{'}} + t y = 7

\int \frac{x ^{2}}{( x ^{2} + 4 ) ^{\frac{3}{2}}} d x

What is the integral of x/(sqrt(x^2+9)) ?
The integral of x/(sqrt(x^2+9)) is sqrt(x^2+9)+C