$$\int\limits_{1}^{3} \frac{2t+3}{t+1}$$
please integrate by definite integration. :)

Question

anonymous · Answer

$$\int\limits_{1}^{3} \frac{2t+3}{t+1}$$

anonymous · Answer

try $u=t+1$

anonymous · Answer

ok. :)

anonymous · Answer

don't let the numerator throw you off, 
if $u=t+1$ then $t=u-1$ and $2t+3)=2(u-1)+3$ etc

blockcolder · Answer

Actually, doing the division makes it simpler:
$$\frac{2t+3}{t+1}=2-\frac{1}{t+1}$$

anonymous · Answer

is answer is

ln(2) + 4  ?

anonymous · Answer

yes! @mrtdmr26 can you show me your solution?

anonymous · Answer

$$[ \int\limits\limits_{1}^{3}(2t)/(t+1) ] + [ \int\limits\limits_{1}^{3} 3 / (t +1) ] $$

$$\int\limits\limits(2t)/(t+1) = [ -2\ln (|t+1|) + 2t - 1 + C]$$

$$ \int\limits\limits 3 / (t +1)  = 3\ln ( |x+1| ) + C$$

when yu put the limits 1 to 3

1st Part  $$4-2\ln(2)$$ 2nd Part$$3\ln(2)$$

$$4-2\ln(2) + 3\ln(2) = \ln(2) + 4$$

That's all :)

anonymous · Answer

Thank you a thousand times @mrtdmr26 . :)