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

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

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integral of (xln(x))/(sqrt(x^2-1))

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

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

Solution steps

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

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int_{1}^{\infty} (\frac{1}{x ^{3}}) d x

\frac{d y}{d x} x + y = - 2 x^{6} y^{4}

4 \frac{d ^{2} x}{d t ^{2}} - 1 \frac{d x}{d t} - 3 x = 0, x (0) = 1

arctan (- 6 x)

4 t^{2} cos (2 t) + 5 t e^{- 7 t}

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