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# Q. `int` `(x^2 - 1)/(x^2 + 1)` `dx`     `=?`

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Q. `int` `(x^2 - 1)/(x^2 + 1)` `dx`     `=?`

Posted by user8235304 on July 12, 2013 at 12:46 AM via web and tagged with calculus, math

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`int(x^2-1)/(x^2+1)dx`

Notice that the degree of the numerator and denominator are the same. So, before we integrate, expand `(x^2-1)/(x^2+1)` .

Using long division,

`(x^2-1)/(x^2+1)=1-2/(x^2+1)`

So,

`int(x^2-1)/(x^2+1)dx`

`=int (1-2/(x^2+1))dx`

Then, express it as difference of two integrals.

`=intdx - int 2/(x^2+1)dx`

For the first integral, apply the formula `int kdx=kx+C` .

`=x+C -int 2/(x^2+1)dx`

For the second integral, apply the formula `int1/(x^2+a^2)dx=1/a tan^(-1) x/a` .

`=x+C -2int 1/(x^2+1)dx`

`=x+C-2*tan^(-1)x+C`

Since C represents any number, express the sum of the C's as C only.

`=x-2tan^(-1)x+C`

Hence, `int(x^2-1)/(x^2+1)dx=x-2tan^(-1)x+C` .

Posted by mjripalda on July 12, 2013 at 2:17 AM (Answer #1)