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How would I go about finding the indefinite integral below? Would I try to change the equation so that I can use the substitution rule? Thanks in advance.

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

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    $\begingroup$ Turn the numerator into differentiation of of denominator by manipulating the constant term in numerator. $\endgroup$ – Mrigank May 18 '17 at 8:44
  • $\begingroup$ You probably also want to complete the square in the denominator to turn whatever remains after following Mrigank's tip into a table form. $\endgroup$ – Jyrki Lahtonen May 18 '17 at 8:45
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    $\begingroup$ BTW. Did you search the site? I am very sure that we have handled integration of rational functions with a positive definite quadratic denominator before. $\endgroup$ – Jyrki Lahtonen May 18 '17 at 8:46
  • $\begingroup$ Same type: math.stackexchange.com/questions/1356419/… $\endgroup$ – k.Vijay May 18 '17 at 8:48
  • $\begingroup$ Another option : sosmath.com/calculus/integration/trigsub/trigsub.html $\endgroup$ – lab bhattacharjee May 18 '17 at 8:53
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Hint:

Rewrite $2x+2$ as $\frac12(4x-2)+3$, so that $$\int \frac{2x+2}{2x^2-2x+1}\,\mathrm dx=\frac12\int \frac{4x-2}{2x^2-2x+1}\,\mathrm dx+3\int \frac{\mathrm dx}{2x^2-2x+1}.$$ Now $$\int \frac{\mathrm dx}{2x^2-2x+1}=2\int \frac{\mathrm dx}{(2x-1)^2+1}$$ and set $u=2x-1$.

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  • $\begingroup$ Are you saying $\int \frac{\mathrm dx}{2x^2-2x+1}$ is equal to $\frac12\int \frac{\mathrm dx}{(2x-1)^2+1}$ ? $\endgroup$ – randb May 18 '17 at 9:21
  • $\begingroup$ Sorry, I messed up with the fractions. We have a factor $2$ (fixed). Thanks for pointing it. $\endgroup$ – Bernard May 18 '17 at 9:35
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$$I = \int \frac{2x+2}{\:2x^2-2x+1}dx$$

Write $$2x+2 = \frac{1}{2}(4x-2) +3$$

With this algebraic manipulation and integral now become:

$$I = \frac{1}{2}\int \frac{4x-2}{\:2x^2-2x+1}dx+\int \frac{3}{\:2x^2-2x+1}dx$$

First solving the first term

Let $$2x^2-2x+1 = u$$ $$(4x-2)dx = du$$

$$I = \frac{1}{2}\int \frac{du}{\:u}+\int \frac{3}{\:2x^2-2x+1}dx$$

$$I = \frac{1}{2}\ln|u|+\int \frac{3}{\:2x^2-2x+1}dx$$

Write

$$\:2x^2-2x+1 = (\sqrt2x - \frac{1}{\sqrt2})^2+\frac{1}{2}$$

$$I = \frac{1}{2}\ln|2x^2-2x+1|+\int \frac{3}{(\sqrt2x - \frac{1}{\sqrt2})^2+\frac{1}{2}}dx$$

Let $$\sqrt2x - \frac{1}{\sqrt2} = \frac{p}{\sqrt2} $$ $$dx = \frac{dp}{2}$$

$$I = \ln\sqrt{|2x^2-2x+1|}+3\int \frac{dp}{p^2+1}dx$$

$$I = \ln\sqrt{|2x^2-2x+1|}+3arctanp + C$$

$$I = \ln\sqrt{|2x^2-2x+1|}+3arctan(2x-1) + C$$

comment on the step you don't understand.

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