How do I find the integral ∫t2t+4dt ?
1 Answer
Sep 30, 2014
=t22−4t−24+16ln(t+4)+c , wherec is a constantExplanation :
=∫t2t+4dt let's
t+4=u , thendt=du
=∫(u−4)2udu
=∫u2−8u+16udu
=∫(u2u−8uu+16u)du
=∫(u−8+16u)du
=∫udu−∫8du+∫16udu
=u22−8u+16lnu+c , wherec is a constantSubstituting
u back yields,
=(t+4)22−8(t+4)+16ln(t+4)+c , wherec is a constantSimplifying further,
=t22+8+4t−8t−32+16ln(t+4)+c , wherec is a constant
=t22−4t−24+16ln(t+4)+c , wherec is a constant