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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{u + 1}{u} : 1 + \frac{1}{u}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 5 \cdot \frac{1}{2} (u + ln ∣ u ∣)

Substitute back

u = x^{2} - 1

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} + 3 y^{^{'}} + 2 y = (1 + e^{t})^{- 1}

4 x^{- 3}

x \to - 5 lim (∣ x ∣ - 5)

(1 - x)^{- 100}

e^{- 0.4 t} cos (12 t)

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