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

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

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

3 (ln ∣ t ∣ - \frac{1}{2} ln t^{2} + 1) + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{1}{t ( 1 + t ^{2} )} d t

Take the partial fraction of

\frac{1}{t ( 1 + t ^{2} )} : \frac{1}{t} - \frac{t}{t ^{2} + 1}

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

Apply the Sum Rule:

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

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

\int \frac{1}{t} d t = ln ∣ t ∣

\int \frac{t}{t ^{2} + 1} d t = \frac{1}{2} ln t^{2} + 1

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

Add a constant to the solution

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

\int \frac{4 x ^{2} - x}{x} d x

\int \frac{1}{g - k v ^{2}} d v

\int \frac{x ^{2}}{x ^{2} + y ^{2}} d x

\int \frac{\partial}{\partial x} (x^{2}) d x

\int tan^{2} (1 - x) d x

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