Evaluate the Integral: $\int\frac{3t-2}{t+1}$
1) Long division gives $\int3-\frac{5}{t+1}dt$
2)I take the negative and constant outside of the integral.
$-3\int\frac{5}{t+1}dt$
3) I then take the 5 out side the integral and multiply it by the three.
$-15\int\frac{1}{t+1}dt$
4) U-substitution
$u=t+1$ and $du=dt$
5) Integrate
$-15\ln|t|+C$
6) Replace $u$ with $t+1$
Final answer is $-15\ln|t+1|+C$
Am I correct if not please explain.
$\endgroup$ 43 Answers
$\begingroup$Mistake begins from step 2.
$$\int3-\frac{5}{t+1}dt \not = -3\int\frac{5}{t+1}dt $$
It should be as follows: $$\int(3-\frac{5}{t+1})dt$$ $$=3\int dt -5\int \frac{d(t+1)}{(t+1)}$$ $$=3t-5\ln\big|t+1\big|+c$$
$\endgroup$ 4 $\begingroup$\begin{align*} \int \left(3-\frac{5}{t+1}\right)\,dt&=\int 3\,dt-\int \frac{5}{t+1}\,dt\\ &=3t-5\ln|t+1|+C \end{align*}
$\endgroup$ 0 $\begingroup$$$\int\frac{3t-2}{t+1}\space\space\text{d}t=$$ $$\int\left(3-\frac{5}{t+1}\right)\space\space\text{d}t=$$ $$-5\int\frac{1}{t+1}\space\space\text{d}t+3\int 1\space\space\text{d}t=$$
Substitute $u=t+1$ and $\text{d}u=\text{d}t$:
$$-5\int\frac{1}{u}\space\space\text{d}u+3\int 1\space\space\text{d}t=$$ $$-5\ln\left(u\right)+3t+\text{C}=$$ $$-5\ln\left(t+1\right)+3t+\text{C}=$$ $$3t-5\ln\left(t+1\right)\text{C}$$
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