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

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

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

integral

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

arctan (x + 1) + C

Solution steps

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

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 = x + 1

= arctan (x + 1)

Add a constant to the solution

= arctan (x + 1) + C

\frac{1}{( a ^{3} - 2 a )}

4 - ∣ x ∣

(9 x + 4)^{4}

cos (2 x + 3)

2 x 2 x - 3

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