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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 5 \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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} - 1

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 5, y^{^{'}} (0) = 0

h \to 0 lim (\frac{h}{h ^{2} - 7 h + 1})

\frac{d}{d x} (10 0.6 x + 7)

f (x) = x^{2} + 40, at x = 3

\frac{d y}{d x} + 7 (tan (7 x)) y = 2 cos (7 x)

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