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How to find the value of $$\int _ { - 5 } ^ { - 2 } \left( \frac { x ^ { 2 } - x } { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx + \int _ { \frac 16 } ^ { \frac 13 } \left( \frac { x ^ { 2 } - x } { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx+\int _ { \frac 65 } ^ {\frac 32 } \left( \frac { x ^ { 2 } - x} { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx$$

My attempt:

I tried to substitute in second and third integral, such that powers of integration become the same. But I could not find such a substitution. tried taking out powers of x from denominator such that differential coefficient gets formed in the numerator .

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  • $\begingroup$ Please use mathjax instead of including a picture of formulas $\endgroup$ – supinf Feb 8 at 10:06
  • $\begingroup$ Hoping that you don't mind, I fixed the edit of the post. Cheers :-) $\endgroup$ – Claude Leibovici Feb 8 at 11:22
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$$I= \int _ { - 5 } ^ { - 2 } \left( \frac { x ^ { 2 } - x } { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx + \int _ { \frac 16 } ^ { \frac 13 } \left( \frac { x ^ { 2 } - x } { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx+\int _ { \frac 65 } ^ {\frac 32 } \left( \frac { x ^ { 2 } - x} { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx$$

$\displaystyle I_{1} = \int^{-2}_{-5}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2dx,$ put $\displaystyle x=1-\frac{1}{u}\rightarrow u=\frac{1}{1-x}$ and $\displaystyle dx = \frac{1}{u^2}du$

$$\displaystyle I_{1}=\int^{\frac{1}{3}}_{\frac{1}{6}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2\frac{1}{x^2}dx$$

$\displaystyle I_{3}=\int^{\frac{3}{2}}_{\frac{6}{5}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2dx,$ put $\displaystyle x=\frac{1}{1-v}\rightarrow v=1-\frac{1}{x}$ and $\displaystyle dx = \frac{1}{(1-v)^2}dv$

$$I_{3} =\int^{\frac{1}{3}}_{\frac{1}{6}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2\frac{1}{(1-x)^2}dx$$

$$I = \int^{\frac{1}{3}}_{\frac{1}{3}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2\bigg[\frac{1}{x^2}+1+\frac{1}{(1-x)^2}\bigg]dx$$

$$I = \int^{\frac{1}{3}}_{\frac{1}{6}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2\bigg[\frac{x^{4}-2x^{3}+3x^{2}-2x+1}{x^2(1-x)^2}\bigg]dx$$

$$I = \int^{\frac{1}{3}}_{\frac{1}{6}}\bigg(\frac{x-x^{2}}{x^{3}-3x+1}\bigg)'dx = \frac{x-x^{2}}{x^{3}-3x+1}\bigg|^{\frac{1}{3}}_{\frac{1}{6}}=$$

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