What is the integral of (7x)/(5x^2-3) ?

The integral of (7x)/(5x^2-3) is 7/10 ln|5x^2-3|+C

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integral of (7x)/(5x^2-3)

\int \frac{7 x}{5 x ^{2} - 3} d x

\frac{7}{10} ln 5 x^{2} - 3 + C

Solution steps

\int \frac{7 x}{5 x ^{2} - 3} d x

Take the constant out:

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

= 7 \cdot \int \frac{x}{5 x ^{2} - 3} d x

Apply u-substitution

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

= 7 \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

= 7 \cdot \frac{1}{10} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 7 \cdot \frac{1}{10} ln ∣ u ∣

Substitute back

u = 5 x^{2} - 3

= 7 \cdot \frac{1}{10} ln 5 x^{2} - 3

Simplify

7 \cdot \frac{1}{10} ln 5 x^{2} - 3 : \frac{7}{10} ln 5 x^{2} - 3

= \frac{7}{10} ln 5 x^{2} - 3

Add a constant to the solution

= \frac{7}{10} ln 5 x^{2} - 3 + C

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What is the integral of (7x)/(5x^2-3) ?
The integral of (7x)/(5x^2-3) is 7/10 ln|5x^2-3|+C