Instructions were evaluate the integral.
(x/1-x^4) dx
I know that this form is the tanh inverse 1/1-x^2 but not sure how to substitute and start this problem. Please Help.
To evaluate the integral ∫(x/(1-x^4)) dx, one possible approach is to use the method of partial fractions.
Let's start by factoring the denominator:
1 - x^4 = (1 - x^2)(1 + x^2)
Now, we can express the fraction as a sum of partial fractions:
x/(1 - x^4) = A/(1 - x^2) + B/(1 + x^2)
To find the values of A and B, we need to solve for them. Let's create a common denominator:
x = A(1 + x^2) + B(1 - x^2)
Expanding and combining like terms:
x = A + Ax^2 + B - Bx^2
We can equate the coefficients of x^2 and the constant term on both sides:
Coefficient of x^2: 0 = A - B
Constant term: x = A + B
From the first equation, we can solve for A: A = B
Substituting into the second equation: x = A + A
2A = x
A = x/2
Since A = B, we have B = x/2 as well.
Now, we can rewrite the integral as:
∫(x/(1 - x^4)) dx = ∫(x/((1 - x^2)(1 + x^2))) dx
= ∫(x/(1 - x^2)) dx + ∫(x/(1 + x^2)) dx
At this point, we can use a trigonometric substitution to simplify each integral. Let's substitute u = x^2 for the first integral (x/(1 - x^2)):
du/dx = 2x
dx = du/(2x)
Substituting back into the first integral, we get:
∫(x/(1 - x^2)) dx = ∫(1/(1 - u)) (du/(2x))
= (1/2) ∫(1/(1 - u)) (du/u)
= (1/2) ln|1 - u| + C
= (1/2) ln|1 - x^2| + C1
For the second integral (x/(1 + x^2)), let another substitution be v = 1 + x^2:
dv/dx = 2x
dx = dv/(2x)
Substituting back into the second integral, we get:
∫(x/(1 + x^2)) dx = ∫(1/v) (dv/(2x))
= (1/2) ∫1/v dv
= (1/2) ln|v| + C
= (1/2) ln|1 + x^2| + C2
Now, combining the two results, we have:
∫(x/(1 - x^4)) dx = (1/2) ln|1 - x^2| + (1/2) ln|1 + x^2| + C
= (1/2) ln|1 - x^2| + ln|√(1 - x^2)| + C
Therefore, the integral evaluates to:
∫(x/(1 - x^4)) dx = (1/2) ln|1 - x^2| + ln|√(1 - x^2)| + C
Note: The natural logarithm (ln) denotes the natural logarithm with base e. The constants C, C1, and C2 represent arbitrary constants of integration.