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

The integral of 1/((s+1)^2+1) is arctan(s+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

integral

\frac{1}{( s + 1 ) ^{2} + 1}

arctan (s + 1) + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

= arctan (u)

Substitute back

u = s + 1

= arctan (s + 1)

Add a constant to the solution

= arctan (s + 1) + C

\frac{1}{( v + 1 ) ^{2}}

3 t^{2} + 2 t + 1

3 t^{2} + 2 t + 2

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

\frac{( a ^{2} - x ^{2} ) ^{3}}{2}

What is the integral of 1/((s+1)^2+1) ?
The integral of 1/((s+1)^2+1) is arctan(s+1)+C