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

f (x) = x^{3}, (0, 0)

f (x) = 3 (5 - x)^{2}

\frac{d}{d x} ((9 x + 2)^{\frac{2}{3}})

\int sin (a x) cos (a x) d x

x \to 0 lim ((\frac{1}{3})^{\frac{1}{x}})

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