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Evaluating an indefinite integral it is simple.

Hello it is simple but I can not. Can you help me? evaluate the following integral:

$$\int \frac{2x+3} {{x^2-2x-3}}\mathrm{d}x$$

Give me show step by step solutions please.

Thank you very much.

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closed as unclear what you're asking by user147263, graydad, Ivo Terek, Mark Viola, Michael Grant Jul 11 '15 at 0:40

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$$\int\frac{2x+3}{x^2-2x-3}$$

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

For integrate $\frac{2x-2}{x^2-2x-3}$ , substitute $t=x^2-2x-3$ and $dt=(2x-2)dx$:

$$=\int\frac{1}{t}dt+5\int\frac{1}{(x-1)^2-4}dx$$

For integrate $\frac{1}{(x-1)^2-4}$ , substitute $z=x-1$ and $dz=dx$:

$$\ln(t)+5\int\frac{1}{z^2-4}dz$$

Factor out $-4$ from the denominator and factor out constant:

$$=\ln(t)+\frac{5}{4}\int\frac{1}{1-\frac{z^2}{4}}dz$$

For integrate $\frac{1}{1-\frac{z^2}{4}}$ substitute $\varphi=\frac{z}{2}$ and $d\varphi=\frac{1}{2}dz$:

$$\ln(t)-\frac{5}{2}\int\frac{1}{1-\varphi ^2}d\varphi$$

The integral of $\frac{1}{1-\varphi ^2}$ is arctanh $(\varphi)$:

$\ln(t)-\frac{5}{2}$ arctanh $(\varphi)+\color{blue}c$

Substitute back for $\varphi=\frac{z}{2}$, $z=x-1$, $t=x^2-2x-3$:

$=\ln(x^2-2x-3)-\frac{5}{2}$ arctanh $(\frac{x-1}{2})+\color{blue}c$

Wich is equivalent for restricted $x$ values to:

$$\boxed{\color{red}{\frac{9}{4}\ln(3-x)-\frac{1}{4}\ln(x+1)}+\color{blue}c}$$

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  • 2
    $\begingroup$ The last line is correct, but how did it come from the other lines? What happened to the arctangent in the previous line? Where did that arctangent come from? Why did you complete the square in the second integral when the denominator factors? $\endgroup$ – Rory Daulton Jul 10 '15 at 14:44
  • $\begingroup$ I can write all the details,it takes more time :) $\endgroup$ – 3SAT Jul 10 '15 at 14:48
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    $\begingroup$ You need to check the derivative of arctanx. This is not the derivative of arctan.how you get it. Somothing wrong $\endgroup$ – Syed Muhammad Asad Jul 10 '15 at 15:23
  • $\begingroup$ Im working on a full answer $\endgroup$ – 3SAT Jul 10 '15 at 15:24
  • $\begingroup$ wtf this is your edit, are you joking with us? $\endgroup$ – Lucas Jul 10 '15 at 15:29
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Hint: write $$\int \frac {2x-2+5}{x^2-2x-3} dx $$ and then integrate two integrals. The first is $\ ln|x^2-2x-3|+k$ and for the second integral ($\int \frac {5}{x^2-2x-3} dx$) use the coefficients $A$ and $B$.

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Hint: $$\int \frac{2x+3} {{x^2-2x-3}}\mathrm{d}x$$

$$=\int \frac{2x-2} {{x^2-2x-3}}\mathrm{d}x+\int \frac{5} {{x^2-2x-3}}\mathrm{d}x$$

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\begin{align} I &= \int \frac{2x+3}{x^2-2x-3} \, dx = \int \frac{2(x-1)+5}{(x-1)^{2} - 4} \, dx \end{align} let $t = x-1$ \begin{align} I &= \int \frac{2 t + 5}{t^{2} - 4} \, dt = \int \frac{2 t + 5}{(t-2)(t+2)} \, dt \\ &= \frac{1}{4} \, \int \left(\frac{2t+5}{t-2} - \frac{2t+5}{t+2}\right) \, dt \\ &= \frac{1}{4} \, \int \left( \frac{9}{t-2} - \frac{1}{t+2} \right) \, dt \\ &= \frac{1}{4} \, \left( 9 \, \ln(t-2) - \ln(t+2) \right) = \frac{1}{4} \, \left( 9 \, \ln(x-3) - \ln(x+1) \right). \end{align}

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As the given $$\int\frac{2x+3}{x^2-2x-3}$$ $$=\int\frac{2x-2}{x^2-2x-3}dx+5\int\frac{1}{x^2-2x-3}dx$$ $$=\ln(x^2-2x-3)+5\int\frac{1}{(x-1)^2-4}$$ $$=\ln(x^2-2x-3)+5\int\frac{1}{(x-1)^2-(2)^2}$$ $$=\ln(x^2-2x-3)+5 \frac{1}{2(2)} \ln \frac{(x-1)-2}{(x-1)+2}$$ $$=\ln(x^2-2x-3)+ \frac{5}{4} \ln \frac{(x-3)}{(x+1)}$$

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There are many ways to do the integral, but the most straightforward way, without tricks, uses partial fractions.

Your denominator is $x^2-2x-3=(x-3)(x+1)$. Therefore we look for constants $A$ and $B$ such that

$$\frac{2x+3}{(x-3)(x+1)}=\frac A{x-3}+\frac B{x+1}$$ $$2x+3=A(x+1)+B(x-3)$$

Substituting $x=-1$ into that gives us $-4B=1$ so $B=-\frac 14$; substituting $x=3$ gives us $4A=9$ so $A=\frac 94$. We can now easily do our integral.

$$\begin{align} \int\frac{2x+3}{(x-3)(x+1)}\,dx &= \int\left(\frac 94\cdot\frac 1{x-3}-\frac 14\cdot\frac 1{x+1}\right)\,dx \\ &= \frac 94\int\frac 1{x-3}\,dx-\frac 14\int\frac 1{x+1}\,dx \\ &= \frac 94\ln|x-3|-\frac 14\ln|x+1|+C \end{align}$$

Note that the absolute value signs make this the most general answer to your integral, valid everywhere except $x=-1$ and $x=3$. As you probably know, the arbitrary constant $C$ could be different in the three intervals $(-\infty,-1), (-1,3),$ and $(3,\infty)$. That includes all the indefinite integrals for your question.

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Hint : computational purpose only

$ \int \frac {px+q}{ax^2+bx+c}dx=\frac{p}{2a}log|{ax^2+bx+c}|+(q-\frac{pb}{2a})log|\frac{2ax+b-\sqrt{\Delta}}{2ax+b+\sqrt{\Delta}}|$

note: use this result only for $b^2-4ac>0$

In your problem 

$a=1,b=-2,c=-3,p=2,q=3,\Delta=16$

$\int \frac {2x+3}{x^2-2x-3}dx$

$=\frac{2}{2(1)}log|x^2-2x-3|+(3-\frac{(2)(-2)}{2(1)})\frac{1}{\sqrt{16}}log|\frac{2x-2-\sqrt{16}}{2x-2+\sqrt{16}}|$

$=1log|x^2-2x-3|+\frac{5}{4}log|\frac{x-3}{x+1}|+c$

Where "c" is integral constant

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  • $\begingroup$ In a quick mood I forget to type log term as first term shown in formulae $\endgroup$ – siva naga kumar Jul 10 '15 at 18:03
  • $\begingroup$ I just edited , you can now check that $\endgroup$ – siva naga kumar Jul 10 '15 at 18:04
  • $\begingroup$ Thank you Rory For your concern $\endgroup$ – siva naga kumar Jul 10 '15 at 18:09
  • $\begingroup$ If you like it please vote for my work $\endgroup$ – siva naga kumar Jul 10 '15 at 18:13

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