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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

ln 1 + t^{2} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 + t^{2}

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

Simplify

2 \cdot \frac{1}{2} ln 1 + t^{2} : ln 1 + t^{2}

= ln 1 + t^{2}

Add a constant to the solution

= ln 1 + t^{2} + C

y = - 2 x^{2} + 20, y = 3 x - 15

f (x) = x^{7} - 1 (x)

\int r 2 - r^{2} d r

\int \frac{1}{( 4 x + 1 ) 16 x ^{2} + 4} d x

f (x) = - 1 - 8 x^{2}, at x = - 2

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