Question

1/x power 4 -1 integrals​

Answer 1

Solution:

Before solving the problem of integration, we must get to know to formulas which we will use in the solution.

1. $\int \frac{dx}{x^{2}+1}=tan^{-1}x+c$

where c is constant of integration.

2. $\int \frac{dx}{x^{2}-1}=\frac{1}{2}log|\frac{x-1}{x+1}|+c$

where c is constant of integration.

Now, we proceed to solve the problem.

Now, $\int \frac{dx}{x^{4}-1}$

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

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

= $\frac{1}{2}\int \frac{dx}{x^{2}-1}-\frac{1}{2}\int \frac{dx}{x^{2}+1}$

= $\frac{1}{2}.\frac{1}{2}log|\frac{x-1}{x+1}|-\frac{1}{2}tan^{-1}x+c$

where c is the constant of integration

= $\frac{1}{4}log|\frac{x-1}{x+1}|-\frac{1}{2}tan^{-1}x+c$

which is the required integral.

Answer 2

Answer:

$\dfrac{1}{4}log\left |\dfrac{x-1}{x+1} \right |-\dfrac{1}{2}tan^-1x+c$

Step-by-step explanation:

In the question:

We have to calculate the integration of the following function:

$\int\frac{1}{x^4-1} dx$

Firstly divide and multiply by 2 to the given function:

$\frac{1}{2} \int \frac{(x^2+1)-(x^2-1)}{x^4-1} dx$

$=\ \dfrac{1}{2} \int\dfrac{x^2+1}{(x^2+1)(x^2-1)} dx-\dfrac{1}{2}\int\dfrac{x^2-1}{(x^2+1)(x^2-1)}dx$

$=\ \dfrac{1}{2} \int\dfrac{1}{x^2-1}dx-\dfrac{1}{2}\int\dfrac{1}{x^2+1}dx$ [by cancel the terms we get this term.]

$=\ \dfrac{1}{4}log\left |\dfrac{x-1}{x+1} \right |-\dfrac{1}{2}tan^-1x+c$

$(1)\ \int\dfrac{1}{x^2-1}dx=\dfrac{1}{2}log\left|\dfrac{x-1}{x+1}\right|+c$

$(2)\ \int\dfrac{1}{x^2+1}dx=tan^-1x+c$

$(3)\ x^4-1=(x^2-1)(x^2+1)$     [formulas used in this question]