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So, here's an example:

$$\int \frac{x^2}{\sqrt{1 - x^2}}\, \mathrm dx$$

My math's teacher just say Add and Subtract 1 like this

$$\int \frac{1 + x^2 - 1}{\sqrt{1 - x^2}}\, \mathrm dx.$$

This is just one example. This thing is in almost half of the problems.

I get the next steps but what disturbs me is WHERE should one multiple/divide or add/subtract a number to make it easier. Of course there must be some concept behind it which I don't know so it will be a great help. I have seen this thing last year while studying trigonometry and this year while studying Calculus. If someone can explain how does this happen. That would be really helpful :)

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  • $\begingroup$ Welcome to MSE. Please read this text about how to ask a good question. $\endgroup$ Apr 23, 2018 at 9:42
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    $\begingroup$ It is experience. Do lots and lots and lots of integration problems and you will see how to re-write expressions in terms of bits you know the integral of. $\endgroup$
    – Paul
    Apr 23, 2018 at 10:13

2 Answers 2

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The whole idea is to simplify fractions, since monomials are easier to integrate than fractions in general. E.g.: $$\int \frac{x}{x+a} \ {\rm d}x =\int \frac{x\color{red}{+a-a}}{x+a} \ {\rm d}x =\int \left(\frac{x+a}{x+a} -\frac{a}{x+a}\right) \ {\rm d}x \\ =\int \left(1 -\frac{a}{x+a}\right) \ {\rm d}x = x-a\ln|x+a|+c$$ rather than using integration by parts with $u=x$ and ${\rm d}v = {\rm d}x/(x+a)$.

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now carry out $$\int\frac{1+x^2-1}{\sqrt{1-x^2}}\ dx=\int\frac{1-(1-x^2)}{\sqrt{1-x^2}}\ dx$$ $$=\int\frac{1}{\sqrt{1-x^2}}\ dx-\int\frac{1-x^2}{\sqrt{1-x^2}}\ dx$$ $$=\int\frac{1}{\sqrt{1-x^2}}\ dx-\int\sqrt{1-x^2}\ dx$$ $$=\sin^{-1}x-\frac12x\sqrt{1-x^2}-\frac12\sin^{-1}x +c$$ $$=\frac12\sin^{-1}x-\frac12x\sqrt{1-x^2}+c$$ so this is final answer

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    $\begingroup$ I think you misunderstood the question... $\endgroup$
    – gebruiker
    Apr 23, 2018 at 13:17
  • $\begingroup$ who down voted my answer? Where am i wrong? $\endgroup$ Apr 23, 2018 at 14:19
  • $\begingroup$ You did not answer the question... $\endgroup$
    – gebruiker
    Apr 24, 2018 at 9:22

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