Question

Evaluate the integral\(\int {\frac{{3t - 2}}{{t + 1}}} dt\).

Short answer

Separate the integral into this form:

\(\int {\frac{{3t}}{{t + 1}}} dt - \int {\frac{2}{{t + 1}}} dt\)

The value of the integral is \(3t - log{\left( {t + 1} \right)^5} + c\)

Step-by-step solution

Separating the integral.

\(\int {\frac{{3t - 2}}{{t + 1}}} dt = \int {\frac{{3t}}{{t + 1}}} dt - \int {\frac{2}{{t + 1}}} dt\)

Solving the individual integrals.

\(\int {\frac{{3t}}{{t + 1}}} = 3\int {\frac{t}{{t + 1}}} \)

Adding and subtracting 1 from the numerator.

\(3\int {\frac{t}{{t + 1}}} = 3\int {\frac{{t + 1 - 1}}{{t + 1}}} = 3\smallint \left( {\frac{{t + 1) - 1}}{{t + 1}}} \right.\)

\( = 3\int {\frac{{t + 1}}{{t + 1}}} - 3\int {\frac{1}{{t + 1}}} \)

\( = 3\int 1 \,dt - 3\int {\frac{1}{{t + 1}}} dt\)

Let\(t + 1 = u\).

\(1 = \frac{{du}}{{dt}} \Rightarrow dt = du.\)

Therefore\(3\int 1 \times dt - 3\int {\frac{1}{{t + 1}}} dt\)

\( = 3\left( t \right) - 3\int {\frac{1}{u}} du\)

\( = 3t - 3log|u| + {c_1}\)

\( = 3t - 3log|t + 1| + {c_1}. \)……..(1)

Now, handling the second integral.

\(\int {\frac{2}{{t + 1}}} dt\)

Let\(t + 1 = y\).

\(1 = \frac{{dy}}{{dt}} \Rightarrow dt = dy.\)

Therefore\(\int {\frac{2}{{t + 1}}} dt = 2\int {\frac{1}{{t + 1}}} dt\).

\( = 2\int {\frac{1}{y}} dy = 2log|y| + {c_2}.\)

\( = 2log|t + 1| + {c_2}\)……(2)

Finding the final answer.

\( = \int {\frac{{3tdt}}{{t + 1}}} - \int {\frac{2}{{t + 1}}} dt\)

\( = 3t - 3log|t + 1| + {c_1} - \left( {2log|t + 1| + {c_2}} \right)\) (Using equations 1 and 2)

\( = 3t - 3log|t + 1| + {c_1} - 2log|t + 1| - {c_2}\)

Rearranging

\( = 3t - 3log|t + 1| - 2log|t + 1| + {c_1} - {c_2}\)

\( = 3t - log{(t + 1)^3} - log{(t + 1)^2} + {c_1} - {c_2}\)

\(\left( {as\,\,n\,logx = log{x^n}} \right)\)

\( = 3t - \left( {log{{(t + 1)}^3} + log{{(t + 1)}^2}} \right) + c\)(as\({c_1} - {c_2}\)is another constant i.e c)

\( = 3t - log\left( {{{(t + 1)}^3}{{(t + 1)}^2}} \right) + c\)\(( as logx + logy = logxy)\)

\( = 3t - log{(t + 1)^5} + c\)

Hence, the value of the integral is \(3t - log{\left( {t + 1} \right)^5} + c\)