How do you find the integral of #int (x-2)/(x^2+1)#?

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Answer 1 · 2017-02-05T15:15:56

#=1/2ln(x^2+1)-(2tan^(-1)x)+C#

We can rewrite the integral as two fractions and then integrate teh em separately.

#int((x-2)/(x^2+1))dx=color(red)(intx/(x^2+1)dx)-color(blue)(int2/(x^2+1)dx#

First integral

#color(red)(intx/(x^2-2)dx)#

using #int(f'(x))/f(x)dx=ln|f(x)|#

#d/(dx)(x^2+1)=2x#

#:.color(red)(intx/(x^2-2)dx=1/2ln|x^2+1|)#

since #x^2+1>0AA in RR " we can omit the modulus sign"#

second integral

#color(blue)(int2/(x^2+1)dx#

we integrate by substitution

#x=tanu=>dx=sec^2udu#

#color(blue)(int2/(x^2+1)dx=2int1/(tan^2u+1)sec^2udu#

#=color(blue)(2(int1/cancel(sec^2u)cancel(sec^2u)du)=2intdu#

#color(blue)(=2u=2tan^(-1)x)#

putting the two results together we have:

#int((x-2)/(x^2+1))dx=color(red)(intx/(x^2+1)dx)-color(blue)(int2/(x^2+1)dx#

#=color(red)(1/2ln(x^2+1))-color(blue)(2tan^(-1)x)+C#