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

The integral of (xln(x^2+1))/(x^2+1) is 1/4 ln^2(x^2+1)+C

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

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

\frac{1}{4} ln^{2} (x^{2} + 1) + C

Solution steps

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

Apply u-substitution

= \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

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

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{ln ^{2} ( x ^{2} + 1 )}{2} : \frac{1}{4} ln^{2} (x^{2} + 1)

= \frac{1}{4} ln^{2} (x^{2} + 1)

Add a constant to the solution

= \frac{1}{4} ln^{2} (x^{2} + 1) + C

\int sec (a) x d x

\frac{\partial}{\partial x} (4 x^{3} + z^{2})

\frac{d}{d x} ((r (x) (x)^{2} - x^{2})^{\frac{1}{2}})

t an g e n t f (x) = sin (3 x), (\frac{π}{18}, \frac{2}{2})

\frac{d}{d x} (sin^{2} (b x) + x^{2} + 1)

What is the integral of (xln(x^2+1))/(x^2+1) ?
The integral of (xln(x^2+1))/(x^2+1) is 1/4 ln^2(x^2+1)+C