Find Indefinite Integral of dx/(x(x^4+1)). I think that Im complicating it too much. I moved the dx out making it 1/(x(x^4+1))dx than 1/(x^5+x)dx. I think i have to use formula indef integral of dx/(x^2+a^2) = 1/a(tan^-1(x/a)) but im stuck i don't know what to do next. I would love some advice! Thnx
To find the indefinite integral of 1/(x(x^4+1)), you are on the right track. You have correctly converted the expression to 1/(x^5+x).
Now, let's see how we can proceed further. The formula you mentioned, which is the integral of 1/(x^2+a^2) is indeed useful here. However, we need to make a certain substitution in order to apply it.
In this case, let's set x^4+1 = u. Taking the derivative of both sides, we get 4x^3 dx = du. Rearranging, we have dx = du / (4x^3).
Now we can rewrite our integral as ∫ 1/(x^5 + x) dx = ∫ 1/(x(x^4 + 1)) dx = ∫ 1/(u * 4x^3) du.
Combining the dx term from above and substituting u, we have ∫ 1/(u * 4x^3) du = 1/4 ∫ 1/(u * x^3) du.
Now we can apply the formula you mentioned for the integral of 1/(x^2+a^2). Here, a = (x^3)^(-1/2) = x^(-3/2).
Using the formula, we have ∫ 1/(u * x^3) du = 1/x^(3/2) * tan^(-1)(u / (x^(-3/2)) + C.
Now substitute the original expression for u, which is x^4 + 1:
∫ 1/(x(x^4+1)) dx = 1/4 * (1/x^(3/2)) * tan^(-1)((x^4+1) / (x^(-3/2))) + C.
Simplifying further, we have:
∫ 1/(x(x^4+1)) dx = 1/4 * (1/x^(3/2)) * tan^(-1)(x^(7/2) + 1/x^(3/2)) + C.
And that is the indefinite integral of dx/(x(x^4+1)).