What is the integral of (3x)/(5x^2-1) ?

The integral of (3x)/(5x^2-1) is 3/10 ln|5x^2-1|+C

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integral of (3x)/(5x^2-1)

\int \frac{3 x}{5 x ^{2} - 1} d x

\frac{3}{10} ln 5 x^{2} - 1 + C

Solution steps

\int \frac{3 x}{5 x ^{2} - 1} 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}{5 x ^{2} - 1} d x

Apply u-substitution

: \int \frac{1}{10 u} d u

= 3 \cdot \int \frac{1}{10 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}{10} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{10} ln ∣ u ∣

Substitute back

u = 5 x^{2} - 1

= 3 \cdot \frac{1}{10} ln 5 x^{2} - 1

Simplify

3 \cdot \frac{1}{10} ln 5 x^{2} - 1 : \frac{3}{10} ln 5 x^{2} - 1

= \frac{3}{10} ln 5 x^{2} - 1

Add a constant to the solution

= \frac{3}{10} ln 5 x^{2} - 1 + C

\int 5 x d x

\frac{d}{d t} (6 e^{t})

cos (x) cos (y) d x + sin (x) sin (y) d y = 0

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

\int (3 x^{3}) d x

What is the integral of (3x)/(5x^2-1) ?
The integral of (3x)/(5x^2-1) is 3/10 ln|5x^2-1|+C