What is the integral of (x^3)/(x^2-25) ?

The integral of (x^3)/(x^2-25) is 1/2 (x^2-25+25ln|x^2-25|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^3)/(x^2-25)

\int \frac{x ^{3}}{x ^{2} - 25} d x

\frac{1}{2} (x^{2} - 25 + 25 ln x^{2} - 25) + C

Solution steps

\int \frac{x ^{3}}{x ^{2} - 25} d x

Apply u-substitution

: \int \frac{u + 25}{2 u} d u

= \int \frac{u + 25}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{u + 25}{u} d u

Expand

\frac{u + 25}{u} : 1 + \frac{25}{u}

= \frac{1}{2} \cdot \int 1 + \frac{25}{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

= \frac{1}{2} (\int 1 d u + \int \frac{25}{u} d u)

\int 1 d u = u

\int \frac{25}{u} d u = 25 ln ∣ u ∣

= \frac{1}{2} (u + 25 ln ∣ u ∣)

Substitute back

u = x^{2} - 25

= \frac{1}{2} (x^{2} - 25 + 25 ln x^{2} - 25)

Add a constant to the solution

= \frac{1}{2} (x^{2} - 25 + 25 ln x^{2} - 25) + C

\frac{\partial}{\partial x} (x^{2} - 2 x + 3)

\frac{d}{d x} (\frac{4 x}{3})

\frac{8}{1 + x ^{4}}, 4 x^{2}

y^{^{'}} + \frac{y}{t} = 3 cos (2 t)

\frac{d y}{d x} = \frac{y + x}{y - x}

What is the integral of (x^3)/(x^2-25) ?
The integral of (x^3)/(x^2-25) is 1/2 (x^2-25+25ln|x^2-25|)+C