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

The integral of (x^2arctan(x))/(1+x^2) is -arctan^2(x)+xarctan(x)+1/2 arctan^2(x)-ln|sqrt(1+x^2)|+C

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

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

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

Solution steps

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

Apply u-substitution

: \int tan^{2} (u) u d u

= \int tan^{2} (u) u d u

Apply Integration By Parts

: u (- u + tan (u)) - \int - u + tan (u) d u

= u (- u + tan (u)) - \int - u + tan (u) d u

\int - u + tan (u) d u = - \frac{u ^{2}}{2} - ln ∣ cos (u) ∣

= u (- u + tan (u)) - (- \frac{u ^{2}}{2} - ln ∣ cos (u) ∣)

Substitute back

u = arctan (x)

= arctan (x) (- arctan (x) + tan (arctan (x))) - (- \frac{arctan ^{2} ( x )}{2} - ln ∣ cos (arctan (x)) ∣)

Simplify

arctan (x) (- arctan (x) + tan (arctan (x))) - (- \frac{arctan ^{2} ( x )}{2} - ln ∣ cos (arctan (x)) ∣) : - arctan^{2} (x) + x arctan (x) + \frac{1}{2} arctan^{2} (x) - ln 1 + x^{2}

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

Add a constant to the solution

= - arctan^{2} (x) + x arctan (x) + \frac{1}{2} arctan^{2} (x) - ln 1 + x^{2} + C

\int \frac{1}{x ^{1.8}} d x

\frac{d y}{d x} - cos (x) y = \frac{1}{100} cos (x)

\int sin (x^{2}) x d x

x \to 2 lim (x - 1)

\frac{d}{d x} (\frac{2 x + 1}{x + 5} (3 x - 1))

What is the integral of (x^2arctan(x))/(1+x^2) ?
The integral of (x^2arctan(x))/(1+x^2) is -arctan^2(x)+xarctan(x)+1/2 arctan^2(x)-ln|sqrt(1+x^2)|+C