Respuesta :
Answer:
[tex]3ln|t+1|+\frac{2}{t+1} +C[/tex]
Explanation:
We'll be using u-substitution for this problem.
Let
[tex]u=t+1\\du=dt[/tex]
Substitute
[tex]\int\limits {\frac{3u-2}{u^2}} \, du[/tex]
Split the fraction
[tex]\int\limits {\frac{3u}{u^2} } \, du -\int\limits {\frac{2}{u^2} } \, du[/tex]
Move the constants out
[tex]3\int\limits {\frac{u}{u^2}du -2\int\limits {u^{-2}} \, du[/tex]
Simplify
[tex]3\int\limits {\frac{1}{u}du -2\int\limits {u^{-2}} \, du[/tex]
Integrate
[tex]3ln|u|+\frac{2}{u} +C[/tex]
Substitute
[tex]3ln|t+1|+\frac{2}{t+1} +C[/tex]
