How do I find the integral intt^2/(t+4)dt ?
1 Answer
Sep 30, 2014
=t^2/2-4t-24+16ln(t+4)+c , wherec is a constantExplanation :
=intt^2/(t+4)dt let's
t+4=u , thendt=du
=int(u-4)^2/udu
=int(u^2-8u+16)/udu
=int(u^2/u-8u/u+16/u)du
=int(u-8+16/u)du
=intudu-int8du+int16/udu
=u^2/2-8u+16lnu+c , wherec is a constantSubstituting
u back yields,
=(t+4)^2/2-8(t+4)+16ln(t+4)+c , wherec is a constantSimplifying further,
=t^2/2+8+4t-8t-32+16ln(t+4)+c , wherec is a constant
=t^2/2-4t-24+16ln(t+4)+c , wherec is a constant