Question #9bf04

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Question #9bf04

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Answer 1 · 2017-06-17T21:47:51

$\frac{- 3 x - 1}{4 (x^{2} + 2 x - 1)} + \frac{3}{4 \sqrt{2}} ln ∣ ∣ ∣ \frac{x - \sqrt{2} + 1}{x + \sqrt{2} + 1} ∣ ∣ ∣ + C$ $(-3x-1)/(4(x^2+2x-1))+3/(4sqrt2)lnabs((x-sqrt2+1)/(x+sqrt2+1))+C$

Explanation:

Rewriting the denominator:

$\int \frac{3 x^{2} + 5 x - 1}{{(x^{2} + 2 x - 1)}^{2}} d x = \int \frac{3 x^{2} + 5 x - 1}{{(x^{2} + 2 x + 1) - 2}^{2}} d x = \int \frac{3 x^{2} + 5 x - 1}{{{(x + 1)}^{2} - 2}^{2}} d x$ $int(3x^2+5x-1)/(x^2+2x-1)^2dx=int(3x^2+5x-1)/{(x^2+2x+1)-2}^2dx=int(3x^2+5x-1)/{(x+1)^2-2}^2dx$

Let $u = x + 1$ $u=x+1$ , implying that $x = u - 1$ $x=u-1$ and $d u = d x$ $du=dx$ .

$= \int \frac{3 {(u - 1)}^{2} + 5 (u - 1) - 1}{{(u^{2} - 2)}^{2}} d u = \int \frac{3 u^{2} - u - 3}{{(u^{2} - 2)}^{2}} d u$ $=int(3(u-1)^2+5(u-1)-1)/(u^2-2)^2du=int(3u^2-u-3)/(u^2-2)^2du$

Now let $u = \sqrt{2} sec θ$ $u=sqrt2sectheta$ . This implies that $u^{2} - 2 = 2 {sec}^{2} θ - 2 = 2 {tan}^{2} θ$ $u^2-2=2sec^2theta-2=2tan^2theta$ and that $d u = \sqrt{2} sec θ tan θ d θ$ $du=sqrt2secthetatanthetad theta$ .

$= \int \frac{3 (2 {sec}^{2} θ) - \sqrt{2} sec θ - 3}{{(2 {tan}^{2} θ)}^{2}} (\sqrt{2} sec θ tan θ d θ)$ $=int(3(2sec^2theta)-sqrt2sectheta-3)/(2tan^2theta)^2(sqrt2secthetatanthetad theta)$

$= \frac{\sqrt{2}}{4} \int \frac{6 {sec}^{3} θ - \sqrt{2} {sec}^{2} θ - 3 sec θ}{{tan}^{3} θ} d θ$ $=sqrt2/4int(6sec^3theta-sqrt2sec^2theta-3sectheta)/tan^3thetad theta$

Multiplying through by $\frac{{cos}^{3} θ}{{cos}^{3} θ}$ $cos^3theta/cos^3theta$ :

$= \frac{1}{2 \sqrt{2}} \int \frac{6 - \sqrt{2} cos θ - 3 {cos}^{2} θ}{{sin}^{3} θ} d θ$ $=1/(2sqrt2)int(6-sqrt2costheta-3cos^2theta)/sin^3thetad theta$

$= \frac{3}{\sqrt{2}} \int {csc}^{3} θ d θ - \frac{1}{2} \int cot θ {csc}^{2} θ d θ - \frac{3}{2 \sqrt{2}} \int \frac{1 - {sin}^{2} θ}{{sin}^{3} θ} d θ$ $=3/sqrt2intcsc^3thetad theta-1/2intcotthetacsc^2thetad theta-3/(2sqrt2)int(1-sin^2theta)/sin^3thetad theta$

Note that $\frac{3}{\sqrt{2}} - \frac{3}{2 \sqrt{2}} = \frac{3}{2 \sqrt{2}}$ $3/sqrt2-3/(2sqrt2)=3/(2sqrt2)$ :

$= \frac{3}{2 \sqrt{2}} \int {csc}^{3} θ d θ - \frac{1}{2} \int cot θ {csc}^{2} θ d θ + \frac{3}{2 \sqrt{2}} \int csc θ d θ$ $=3/(2sqrt2)intcsc^3thetad theta-1/2intcotthetacsc^2thetad theta+3/(2sqrt2)intcscthetad theta$

Find the method for integrating ${csc}^{3} θ$ $csc^3theta$ here.

For the second integral, let $v = cot θ$ $v=cottheta$ so $d v = - {csc}^{2} θ d θ$ $dv=-csc^2thetad theta$ .

The integral of $csc θ$ $csctheta$ is well known. For its derivation, see here.

$= \frac{3}{2 \sqrt{2}} (- \frac{1}{2}) (cot θ csc θ + ln | cot θ + csc θ |) + \frac{1}{2} \int v d v + \frac{3}{2 \sqrt{2}} ln | csc θ - cot θ |$ $=3/(2sqrt2)(-1/2)(cotthetacsctheta+lnabs(cottheta+csctheta))+1/2intvdv+3/(2sqrt2)lnabs(csctheta-cottheta)$

Note that $\frac{1}{2} \int v d v = \frac{1}{2} (\frac{v^{2}}{2}) = \frac{v^{2}}{4} = \frac{{cot}^{2} θ}{4}$ $1/2intvdv=1/2(v^2/2)=v^2/4=cot^2theta/4$ .

$= - \frac{3}{4 \sqrt{2}} cot θ csc θ - \frac{3}{4 \sqrt{2}} ln | cot θ + csc θ | + \frac{3}{4 \sqrt{2}} ln | csc θ - cot θ | + \frac{1}{4} {cot}^{2} θ$ $=-3/(4sqrt2)cotthetacsctheta-3/(4sqrt2)lnabs(cottheta+csctheta)+3/(4sqrt2)lnabs(csctheta-cottheta)+1/4cot^2theta$

Our substitution was $u = \sqrt{2} sec θ$ $u=sqrt2sectheta$ , implying that $cos θ = \frac{\sqrt{2}}{u}$ $costheta=sqrt2/u$ , which is a right triangle where the side adjacent to $θ$ $theta$ is $\sqrt{2}$ $sqrt2$ , the hypotenuse is $u$ $u$ , and the side opposite $θ$ $theta$ is $\sqrt{u^{2} - 2}$ $sqrt(u^2-2)$ .

Thus, $csc θ = \frac{u}{\sqrt{u^{2} - 2}}$ $csctheta=u/sqrt(u^2-2)$ and $cot θ = \frac{\sqrt{2}}{\sqrt{u^{2} - 2}}$ $cottheta=sqrt2/sqrt(u^2-2)$ .

Also note that $- \frac{3}{4 \sqrt{2}} ln | cot θ + csc θ | + \frac{3}{4 \sqrt{2}} ln | csc θ - cot θ | = \frac{3}{4 \sqrt{2}} ln ∣ ∣ ∣ \frac{csc θ - cot θ}{cot θ + csc θ} ∣ ∣ ∣ = \frac{3}{4 \sqrt{2}} ln ∣ ∣ ∣ \frac{1 - cos θ}{1 + cos θ} ∣ ∣ ∣$ $-3/(4sqrt2)lnabs(cottheta+csctheta)+3/(4sqrt2)lnabs(csctheta-cottheta)=3/(4sqrt2)lnabs((csctheta-cottheta)/(cottheta+csctheta))=3/(4sqrt2)lnabs((1-costheta)/(1+costheta))$ .

$= - \frac{3}{4 \sqrt{2}} \frac{\sqrt{2} u}{u^{2} - 2} + \frac{3}{4 \sqrt{2}} ln ∣ ∣ ∣ ∣ \frac{1 - \frac{\sqrt{2}}{u}}{1 + \frac{\sqrt{2}}{u}} ∣ ∣ ∣ ∣ + \frac{1}{4} (\frac{2}{u^{2} - 2})$ $=-3/(4sqrt2)(sqrt2u)/(u^2-2)+3/(4sqrt2)lnabs((1-sqrt2/u)/(1+sqrt2/u))+1/4(2/(u^2-2))$

$= - \frac{3}{4} (\frac{u}{u^{2} - 2}) + \frac{1}{2} (\frac{1}{u^{2} - 2}) + \frac{3}{4 \sqrt{2}} ln ∣ ∣ ∣ \frac{u - \sqrt{2}}{u + \sqrt{2}} ∣ ∣ ∣$ $=-3/4(u/(u^2-2))+1/2(1/(u^2-2))+3/(4sqrt2)lnabs((u-sqrt2)/(u+sqrt2))$

$= \frac{- 3 u + 2}{4 (u^{2} - 2)} + \frac{3}{4 \sqrt{2}} ln ∣ ∣ ∣ \frac{u - \sqrt{2}}{u + \sqrt{2}} ∣ ∣ ∣$ $=(-3u+2)/(4(u^2-2))+3/(4sqrt2)lnabs((u-sqrt2)/(u+sqrt2))$

With $u = x + 1$ $u=x+1$ :

$= \frac{- 3 x - 1}{4 (x^{2} + 2 x - 1)} + \frac{3}{4 \sqrt{2}} ln ∣ ∣ ∣ \frac{x - \sqrt{2} + 1}{x + \sqrt{2} + 1} ∣ ∣ ∣ + C$ $=(-3x-1)/(4(x^2+2x-1))+3/(4sqrt2)lnabs((x-sqrt2+1)/(x+sqrt2+1))+C$

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Question #9bf04

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