sushichan*-* Posted November 20, 2014 Report Share Posted November 20, 2014 Today I saw a integrating question and got paralysed. Where do I even start with this question?? How do I integrate this equation with respect to x:x3/(x-2)3 Reply Link to post Share on other sites More sharing options...
Vioh Posted November 20, 2014 Report Share Posted November 20, 2014 Today I saw a integrating question and got paralysed. Where do I even start with this question?? How do I integrate this equation with respect to x: x3/(x-2)3 First go through all the integration methods that you have learnt from before. Integration by parts? nah, cuz the integral is too complicated, and thus requires you to integrate it by parts several times. And it's hard to believe that it'll work. You also try to use integration by partial fractions, which seems to be very tempting, but since it's not within the syllabus, there's no need to even consider it. The only method left is substitution. Here you can try to substitute u=x-2 --> x=u+2 --> dx=du Thus you basically have to solve the integral (u+2)^3 / u^3 with respect to 'u'. Now, you just have to expand (u+2)^3 using binomial theorem, and then split the integrals to solve for each part individually. The answer you should then get is u + 6ln(u) - 12/u - 4/u^2 = x + 6ln(x-2) - 12/(x-2) - 4/(x-2)^2 + c (the -2 is basically merged into the constant) If you don't understand the symbols that i've written here, just tell me, and i'll rewrite it in the neat Unicode format 3 Reply Link to post Share on other sites More sharing options...
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