What is the integral of ((3+u))/(-(u^2-1)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((3+u))/(-(u^2-1))

\int \frac{( 3 + u )}{- ( u ^{2} - 1 )} d u

\frac{3}{2} ln ∣ u + 1 ∣ - \frac{3}{2} ln ∣ u - 1 ∣ - \frac{1}{2} ln u^{2} - 1 + C

Solution steps

\int \frac{3 + u}{- ( u ^{2} - 1 )} d u

Take the constant out:

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

= - \int \frac{3 + u}{u ^{2} - 1} d u

Expand

\frac{3 + u}{u ^{2} - 1} : \frac{3}{u ^{2} - 1} + \frac{u}{u ^{2} - 1}

Apply the Sum Rule:

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

= - (\int \frac{3}{u ^{2} - 1} d u + \int \frac{u}{u ^{2} - 1} d u)

\int \frac{3}{u ^{2} - 1} d u = - 3 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣)

\int \frac{u}{u ^{2} - 1} d u = \frac{1}{2} ln u^{2} - 1

= - (- 3 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣) + \frac{1}{2} ln u^{2} - 1)

Simplify

- (- 3 (\frac{1}{2} ln ∣ u + 1 ∣ - \frac{1}{2} ln ∣ u - 1 ∣) + \frac{1}{2} ln u^{2} - 1) : \frac{3}{2} ln ∣ u + 1 ∣ - \frac{3}{2} ln ∣ u - 1 ∣ - \frac{1}{2} ln u^{2} - 1

= \frac{3}{2} ln ∣ u + 1 ∣ - \frac{3}{2} ln ∣ u - 1 ∣ - \frac{1}{2} ln u^{2} - 1

Add a constant to the solution

= \frac{3}{2} ln ∣ u + 1 ∣ - \frac{3}{2} ln ∣ u - 1 ∣ - \frac{1}{2} ln u^{2} - 1 + C

y^{^{'}} = \frac{( 6 t ^{2} )}{y}

x \to a lim (- 6)

- 0.5 cos (8 t)

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + y = \frac{e ^{x}}{x ^{2} + 1}

\frac{d}{d x} ((x)^{\frac{3}{2}})

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