How to evaluate the following indefinite integral? $\int\frac{1}{x(x^2-1)}dx.$

Question

I need the step by step solution of this integral please help me! I can't solve it!

$$\int\frac{1}{x(x^2-1)}dx.$$

Answer 1

We use partial fraction decomposition:

$$\int\frac{1}{x(x^2-1)}dx = \int \frac 1{x(x-1)(x+1)}\,dx = \int \left(\frac A{x} + \frac{B}{x - 1} + \frac C{x+1}\right)\,dx$$

Solving for $A, B, C$:

$$A(x-1)(x+1) + Bx(x+1) + Cx(x-1) = 1$$

When $x = 1 \implies 2B = 1 \implies B = \frac 12$

$x = -1 \implies 2C = 1 \iff C = \frac 12$

$x = 0 \implies -A = 1 \iff A = -1$.

That gives us: $$\int \left(\frac {-1}{x} + \frac{1}{2(x - 1)} + \frac 1{2(x+1)}\right)\,dx$$

Now use the fact that $\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$.

Answer 2

Hint:
Use partial fraction decomposition to prove that: $$\dfrac1{x(x^2-1)}=-\dfrac{1}{x}+\dfrac{1}{2(x+1)}+\dfrac1{2(x-1)}.$$ The rest is straightforward.

Answer 3

$$I=\int\frac{dx}{x(x^2-1)}=\int\frac{x\ dx}{x^2(x^2-1)}$$

Setting $x^2=y,2x\ dx=dy$

$$2I=\int\frac{dy}{y(y-1)}=\int\frac{\{y-(y-1)\}dy}{y(y-1)}=\int\frac{dy}{y-1}-\int\frac{dy}y$$ $$=\ln|y-1|-\ln |y|+K$$

$$=\ln|x^2-1|-\ln |x^2|+K$$

$$2I=\ln|x^2-1|-2\ln |x|+K$$

Answer 4

Partial fractions always works. However, it looks like the fastest way might be to multiply top and bottom by $x^{-3}$.

$$\int\frac{dx}{x(x^2-1)}=\int\frac{x^{-3}dx}{xx^{-1}[x^{-2}(x^2-1)]}=\frac12\int\frac{2x^{-3}dx}{1-x^{-2}}=\frac12\ln|1-x^{-2}|+C$$

Answer 5

1/{x*(x^2-1)} =x/{x^2*(x^2-1)} If we substitute: x^2=z By differentiating both sides 2x dx = dz x dx= dz/2 Now if we solve the integral (1/2)log{(x^2-1)/x^2}+C

Answer 6

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