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

The integral of t/(t^2+1) is 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 t/(t^2+1)

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

\frac{1}{2} ln t^{2} + 1 + C

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = t^{2} + 1

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

Add a constant to the solution

= \frac{1}{2} ln t^{2} + 1 + C

n = 1 \sum \infty (n + 1) x^{n}

\frac{ln ( x )}{x ^{2}}

x \to 0 + lim (\frac{2 ^{x}}{3 ^{x}})

\int \frac{( x + 1 )}{x ^{2} + 2 x + 8} d x

\int x^{3} e^{x^{4 + 5}} d x

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