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

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

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

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

arctan^{2} (n) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{arctan ( n )}{1 + n ^{2}} d n

Apply u-substitution

: \int u d u

= 2 \cdot \int u d u

Apply the Power Rule

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

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

Substitute back

u = arctan (n)

= 2 \cdot \frac{arctan ^{2} ( n )}{2}

Simplify

2 \cdot \frac{arctan ^{2} ( n )}{2} : arctan^{2} (n)

= arctan^{2} (n)

Add a constant to the solution

= arctan^{2} (n) + C

\int_{6}^{7} \frac{x}{x ^{2} + 4 x + 20} d x

\int \frac{x ^{3}}{x ^{2} + 2} d x

\int \frac{cos ( a x )}{1 - sin ( a x )} d x

n = 1 \sum \infty \frac{x ^{n}}{4 n - 1}

y = e^{(4 x + 3)}

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