How do you evaluate the integral of $\int ln (x^{2} - 1) d x$ $intln(x^2 - 1)dx$ ?

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Answer 1 · 2016-02-18T13:16:21

Use integration by parts and integration by partial fractions to find that the integral evaluates to

$x ln (x^{2} - 1) - 2 x - ln | x - 1 | + ln | x + 1 | + C$ $xln(x^2-1)-2x-ln|x-1|+ln|x+1|+C$

We start by using integration by parts.

Let $u = ln (x^{2} - 1)$ $u = ln(x^2-1)$ and $d v = d x$ $dv = dx$

Then $d u = \frac{2 x}{x^{2} - 1} d x$ $du = (2x)/(x^2-1)dx$ and $v = x$ $v = x$

By the integration by parts formula $\int u d v = u v - \int v d u$ $intudv = uv - intvdu$

$\int ln (x^{2} - 1) d x = x ln (x^{2} - 1) - \int \frac{2 x^{2}}{x^{2} - 1} d x$ $intln(x^2-1)dx = xln(x^2-1) - int(2x^2)/(x^2-1)dx$

$= x ln (x^{2} - 1) - 2 \int (1 + \frac{1}{x^{2} - 1}) d x$ $=xln(x^2-1)-2int(1+1/(x^2-1))dx$

$= x ln (x^{2} - 1) - 2 \int 1 d x - 2 \int \frac{1}{x^{2} - 1} d x$ $=xln(x^2-1)-2int1dx -2int1/(x^2-1)dx$

$= x ln (x^{2} - 1) - 2 x - 2 \int \frac{1}{(x + 1) (x - 1)} d x$ $=xln(x^2-1)-2x-2int1/((x+1)(x-1))dx$

To evaluate the remaining integral, we can use partial fractions.

$\frac{1}{(x + 1) (x - 1)} = \frac{A}{x + 1} + \frac{B}{x - 1}$ $1/((x+1)(x-1))=A/(x+1)+B/(x-1)$

$\Rightarrow 1 = A (x - 1) + B (x + 1)$ $=> 1 = A(x-1) + B(x+1)$

$= (A + B) x + (- A + B)$ $=(A+B)x + (-A+B)$

$=>{(A+B=0), (-A+B=1):}$

$=>{(A = -1/2),(B=1/2):}$

$=>int(1/((x+1)(x-1))dx = int((1/2)/(x-1)-(1/2)/(x+1))dx$

$=1/2(int1/(x-1)dx-int1/(x+1)dx)$

$=1/2(ln|x-1|-ln|x+1|)+C$

Substituting this back into the prior result, we have

$intln(x^2-1)dx =$
$= xln(x^2-1)-2x-ln|x-1|+ln|x+1|+C$

How do you evaluate the integral of intln(x^2 - 1)dx∫ln(x2−1)dxintln(x^2 - 1)dx?