Evaluating Indefinite Integral, $\int \frac{2x+2}{\:2x^2-2x+1}dx$

Question

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.

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

Answer 1

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 $u=2x-1$.

Answer 2

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

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.