Question #ff170

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Question #ff170

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Answer 1 · 2017-02-21T23:29:05

$3 ln (x^{2} + 2 x + 2) - 12 {tan}^{- 1} (x + 1) + C$ $3ln(x^2+2x+2)-12tan^-1(x+1)+C$

Explanation:

$\int \frac{6 x - 6}{x^{2} + 2 x + 2} d x$ $int(6x-6)/(x^2+2x+2)dx$

$= \int \frac{6 x - 6}{(x^{2} + 2 x + 1) + 1} d x$ $=int(6x-6)/((x^2+2x+1)+1)dx$

$= \int \frac{6 x - 6}{{(x + 1)}^{2} + 1} d x$ $=int(6x-6)/((x+1)^2+1)dx$

In order to make the denominator resemble the trig identity ${tan}^{2} θ + 1 = {sec}^{2} θ$ $tan^2theta+1=sec^2theta$ , let $tan θ = x + 1$ $tantheta=x+1$ .

This also imply that ${sec}^{2} θ . d θ = d x$ $sec^2thetacolor(white).d theta=dx$ and $x = tan θ - 1$ $x=tantheta-1$ . Substituting these in gives:

$= \int \frac{6 (tan θ - 1) - 6}{{tan}^{2} θ + 1} ({sec}^{2} θ . d θ)$ $=int(6(tantheta-1)-6)/(tan^2theta+1)(sec^2thetacolor(white).d theta)$

$= \int \frac{6 tan θ - 12}{{sec}^{2} θ} ({sec}^{2} θ . d θ)$ $=int(6tantheta-12)/sec^2theta(sec^2thetacolor(white).d theta)$

$= \int (6 tan θ - 12) d θ$ $=int(6tantheta-12)d theta$

Both of these are standard integrals. If you forget the integration of $tan θ$ $tantheta$ , recall that $\int tan θ . d θ = \int \frac{sin θ}{cos θ} d θ$ $inttanthetacolor(white).d theta=intsintheta/costhetad theta$ , then use the substitution $u = cos θ$ $u=costheta$ .

$= - 6 ln | cos θ | - 12 θ + C$ $=-6lnabscostheta-12theta+C$

Our original was substitution was $tan θ = x + 1$ $tantheta=x+1$ , so $θ = {tan}^{- 1} (x + 1)$ $theta=tan^-1(x+1)$ .

Furthermore, if $tan θ = x + 1$ $tantheta=x+1$ , this is a triangle where the side opposite $θ$ $theta$ is $x + 1$ $x+1$ and the side adjacent is $1$ $1$ , so the hypotenuse is $\sqrt{{(x + 1)}^{2} + 1} = \sqrt{x^{2} + 2 x + 2}$ $sqrt((x+1)^2+1)=sqrt(x^2+2x+2)$ .

Then, $cos θ = \frac{adjacent}{hypotenuse} = \frac{1}{\sqrt{x^{2} + 2 x + 2}}$ $costheta="adjacent"/"hypotenuse"=1/sqrt(x^2+2x+2)$ . So:

$= - 6 ln ∣ ∣ ∣ \frac{1}{\sqrt{x^{2} + 2 x + 2}} ∣ ∣ ∣ - 12 {tan}^{- 1} (x + 1)$ $=-6lnabs(1/sqrt(x^2+2x+2))-12tan^-1(x+1)$

Using $\frac{1}{\sqrt{x^{2} + 2 x + 2}} = {(x^{2} + 2 x + 2)}^{- \frac{1}{2}}$ $1/sqrt(x^2+2x+2)=(x^2+2x+2)^(-1/2)$ and the log rule $ln (a^{b}) = b ln (a)$ $ln(a^b)=bln(a)$ , this becomes:

$= 3 ln (x^{2} + 2 x + 2) - 12 {tan}^{- 1} (x + 1) + C$ $= color(blue)(3ln(x^2+2x+2)-12tan^-1(x+1)+C$

The absolute value bars aren't necessary because $x^{2} + 2 x + 2 > 0$ $x^2+2x+2>0$ for all real values of $x$ $x$ .

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Question #ff170

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