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# Help me with this integral :(

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$$\displaystyle\int\dfrac{dx}{x^3(1+x^2)}$$

For those who don't use LaTeX:

int(1/((x^3)(1+x^2)) dx)

Jul 27, 2017
edited by MaxWong  Jul 27, 2017

#1
+1

Take the integral:
integral1/(x^3 (x^2 + 1)) dx
For the integrand 1/(x^3 (x^2 + 1)), substitute u = x^2 and du = 2 x dx:
= 1/2 integral1/(u^2 (u + 1)) du
For the integrand 1/(u^2 (u + 1)), use partial fractions:
= 1/2 integral(1/u^2 + 1/(u + 1) - 1/u) du
Integrate the sum term by term and factor out constants:
= 1/2 integral1/(u + 1) du - 1/2 integral1/u du + 1/2 integral1/u^2 du
For the integrand 1/(u + 1), substitute s = u + 1 and ds = du:
= 1/2 integral1/s ds - 1/2 integral1/u du + 1/2 integral1/u^2 du
The integral of 1/s is log(s):
= (log(s))/2 - 1/2 integral1/u du + 1/2 integral1/u^2 du
The integral of 1/u is log(u):
= (log(s))/2 - (log(u))/2 + 1/2 integral1/u^2 du
The integral of 1/u^2 is -1/u:
= (log(s))/2 - 1/(2 u) - (log(u))/2 + constant
Substitute back for s = u + 1:
= -(u log(u) - u log(u + 1) + 1)/(2 u) + constant
Substitute back for u = x^2:
= 1/2 (-1/x^2 - log(x^2) + log(x^2 + 1)) + constant
An alternative form of the integral is:
= 1/2 (log(1/x^2 + 1) - 1/x^2) + constant
Which is equivalent for restricted x values to:
Answer: | = 1/2 (-1/x^2 + ln(x^2 + 1) - 2 ln(x)) + constant

P.S. Whenever you see "log" it means "ln".

Jul 27, 2017
edited by Guest  Jul 27, 2017
#2
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Split into partial fractions, $$\displaystyle \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x^{3}}+\frac{Dx+E}{1+x^{2}}$$ , and integrate term by term.

The fourth term would split to get you a log plus an inverse tangent, (though actually it turns out that $$\displaystyle \text{E = 0}$$ ).

($$\displaystyle \text{ A = -1, C = 1, D = 1, B = E = 0}$$ ).

Tiggsy

Jul 27, 2017
#3
+27318
+2

Can be done as follows:

.Need to add a constant of course.

Jul 27, 2017