compute the integral: (8x^2+6)/(x^2+1)(x+7)

Question

compute the integral:  (8x^2+6)/(x^2+1)(x+7)

anonymous · Answer

It is basically 
$$\int\limits_{}^{} 8x ^{2}+6 - \int\limits_{}^{}x ^{3}+7x ^{2}+x+7$$

anonymous · Answer

thank you so much

anonymous · Answer

i also have another question :( 
compute the indefinite integral: dx/(x^2+4)^(5/2)

anonymous · Answer

i am sorry, I was thinking of logarithms. Please ignore my previous answer. My integration is rusty. i'll get back to you with the correct answer.

anonymous · Answer

okay

anonymous · Answer

the original question is; compute the integral:$$\int\limits_0{}^{1} (8x^2+6) / (x^2+1)(x+7) dx $$

anonymous · Answer

do you know integration by partial fractions? Understanding of that is essential to this problem http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/PartialFrac.html scroll down till you see how the partial fractions are done

anonymous · Answer

thanks !

anonymous · Answer

This is how to solve your problem. The numerator in this example is also a quadratic equation and the denominator is a cubic equation: http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracsoldirectory/PartialFracSol2.html#SOLUTION%209

anonymous · Answer

got it:)

anonymous · Answer

anonymous · Answer