Solve ∫ wrt x/x^2-2x-3dx | Microsoft Math Solver

\int \frac { d x } { x ^ { 2 } - 2 x - 3 } d x

Evaluate

dx\times \frac{\ln(|x-3|)-\ln(|x+1|)}{4}+Сdx

Differentiate w.r.t. x

\frac{d\left(-x^{2}\ln(|x+1|)+x^{2}\ln(|x-3|)+2x\ln(|x+1|)-2x\ln(|x-3|)+3\ln(|x+1|)-3\ln(|x-3|)+4x\right)}{4\left(x-3\right)\left(x+1\right)}+\frac{С\left(dx^{2}-2dx-3d\right)}{x^{2}-2x-3}

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Integration

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\int \frac { d x } { x ^ { 2 } - 2 x - 3 } d x

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How do you evaluate the integral \displaystyle\int\frac{{1}}{{{x}^{{2}}-{2}{x}-{3}}}{\left.{d}{x}\right.} from 0 to 4?

https://socratic.org/questions/how-do-you-evaluate-the-integral-int-1-x-2-2x-3-dx-from-0-to-4

We seek: \displaystyle{\int_{{0}}^{{4}}}\ \frac{{1}}{{{x}^{{2}}-{2}{x}-{3}}}\ {\left.{d}{x}\right.}=-\frac{{1}}{{4}}\ {\left({\ln{{5}}}+{\ln{{3}}}\right)} Explanation: We seek: \displaystyle{I}={\int_{{0}}^{{4}}}\ \frac{{1}}{{{x}^{{2}}-{2}{x}-{3}}}\ {\left.{d}{x}\right.} ...

Evaluating the indefinite integral \int \frac{2x+3} {{x^2-2x-3}}\mathrm{d}x [closed]

https://math.stackexchange.com/q/1356419

\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 ...

Evaluating \int \frac{dx}{x^2 - 2x} dx

https://math.stackexchange.com/questions/156566/evaluating-int-fracdxx2-2x-dx

My book is telling me that I have to complete the square I=\begin{eqnarray*} \int \frac{dx}{x^{2}-2x} &=&\int \frac{dx}{\left( x-1\right) ^{2}-1}\overset{ u=x-1}{=}\int \frac{1}{u^{2}-1}\,du=-\text{arctanh }u+C \end{eqnarray*}, ...

The Integral \int \frac {dx}{(x^2-2ax+b)^n}

https://math.stackexchange.com/questions/3086878/the-integral-int-frac-dxx2-2axbn

All right, now I've got it. The easiest way to get all the coefficients? Expand in a Laurent series around one of the roots. Substituting z=x-a-\sqrt{\Delta} and later defining D=\frac1{2\sqrt{\Delta}} ...

How do you evaluate \displaystyle\int\frac{{\left.{d}{x}\right.}}{{{x}^{{2}}-{2}{x}+{2}}} from \displaystyle{\left[{0},{2}\right]} ?

https://socratic.org/questions/how-do-you-evaluate-int-dx-x-2-2x-2-from-0-2

\displaystyle\frac{\pi}{{2}} Explanation: The denominator cannot be factored, so complete the square and see what you can do in the way of u-substitution. \displaystyle{\int_{{0}}^{{2}}}\frac{{1}}{{{1}{\left({x}^{{2}}-{2}{x}+{1}-{1}\right)}+{2}}}{\left.{d}{x}\right.} ...

Can anyone help me, please? \displaystyle\int\frac{{\left.{d}{x}\right.}}{{{x}^{{2}}-{4}{x}-{13}}}

https://socratic.org/questions/can-anyone-help-me-please

\displaystyle\int\ \frac{{1}}{{{x}^{{2}}-{4}{x}-{13}}}\ {\left.{d}{x}\right.}=\frac{{1}}{{{2}\sqrt{{{17}}}}}\ {\left\lbrace{\ln}{\left|{x}-{2}-\sqrt{{{17}}}\right|}-{\ln}{\left|{x}-{2}+\sqrt{{{17}}}\right|}\right\rbrace}+{C} ...

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