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

7 t \cdot \frac{d y}{d t} + 3 y = t

\int \frac{2 e ^{2 x}}{e ^{2 x} + 14 e ^{x} + 48} d x

\int t^{3} 9 t^{4} + 1 d t

\frac{d y}{d x} = 3 x^{2} (1 + y^{2})

y^{^{'}} = \frac{4 x sec ( \frac{y}{x} ) + y}{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