What is the integral of (3x)/(sqrt(9+x^2)) ?

The integral of (3x)/(sqrt(9+x^2)) is 3sqrt(9+x^2)+C

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integral of (3x)/(sqrt(9+x^2))

\int \frac{3 x}{9 + x ^{2}} d x

3 9 + x^{2} + C

Solution steps

\int \frac{3 x}{9 + x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{9 + x ^{2}} d x

Apply u-substitution

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

= 3 \cdot \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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= 3 \cdot \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 9 + x^{2}

= 3 \cdot \frac{1}{2} \cdot 2 (9 + x^{2})^{\frac{1}{2}}

Simplify

3 \cdot \frac{1}{2} \cdot 2 (9 + x^{2})^{\frac{1}{2}} : 3 9 + x^{2}

= 3 9 + x^{2}

Add a constant to the solution

= 3 9 + x^{2} + C

f (x) = 2^{2 x}

\int_{1}^{5} 2 r^{2} ln (r) d r

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

y = x, y = 2 - x, y = 0

y = 2 sin (x), y = e^{x}, x = 0, x = \frac{π}{2}

What is the integral of (3x)/(sqrt(9+x^2)) ?
The integral of (3x)/(sqrt(9+x^2)) is 3sqrt(9+x^2)+C