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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{x ln ( x ^{2} + 1 )}{x ^{2} + 1} d x

Apply u-substitution

: \int \frac{ln ( u )}{2 u} d u

= 2 \cdot \int \frac{ln ( u )}{2 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{ln ( u )}{u} d u

Apply u-substitution

: \int v d v

= 2 \cdot \frac{1}{2} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= 2 \cdot \frac{1}{2} \cdot \frac{v ^{2}}{2}

Substitute back

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

Simplify

2 \cdot \frac{1}{2} \cdot \frac{ln ^{2} ( x ^{2} + 1 )}{2} : \frac{1}{2} ln^{2} (x^{2} + 1)

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

Add a constant to the solution

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

x^{2} + 16, at x = 3

\int \frac{1}{3} x^{3} 9 - x^{2} d x

\frac{d}{d x} (2 x^{2} + 5)

y = cot^{2} (cos (x))

\frac{d}{d t} (\frac{t}{t - 2})

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