What is the integral of (3x)/(x^2+5) ?

The integral of (3x)/(x^2+5) is 3/2 ln|x^2+5|+C

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integral of (3x)/(x^2+5)

\int \frac{3 x}{x ^{2} + 5} d x

\frac{3}{2} ln x^{2} + 5 + C

Solution steps

\int \frac{3 x}{x ^{2} + 5} 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}{x ^{2} + 5} 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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 5

= 3 \cdot \frac{1}{2} ln x^{2} + 5

Simplify

3 \cdot \frac{1}{2} ln x^{2} + 5 : \frac{3}{2} ln x^{2} + 5

= \frac{3}{2} ln x^{2} + 5

Add a constant to the solution

= \frac{3}{2} ln x^{2} + 5 + C

4 y^{^{'}} + t y = 3

\int x^{5} e^{2 x} d x

\int csc^{2} (5 x) d x

\int_{1}^{7} (t^{2} - 4 t - 12) d t

x \to 4 - lim (1 + 4 - x)

What is the integral of (3x)/(x^2+5) ?
The integral of (3x)/(x^2+5) is 3/2 ln|x^2+5|+C