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

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

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

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

\frac{arctan ^{2} ( x )}{2} + C

Solution steps

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

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{u ^{2}}{2}

Substitute back

u = arctan (x)

= \frac{arctan ^{2} ( x )}{2}

Add a constant to the solution

= \frac{arctan ^{2} ( x )}{2} + C

\int 36 - 4 x^{2} d x

\int 3 x e^{- 3 x} d x

y = x^{2} - 2 x, y = 3 x + 6

\int e^{- x} sin (4 x) d x

(x^{2} + 1) y^{^{'}} + 6 x (y - 1) = 0, y (0) = 8

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