Question

Evaluate the integral.

\(\int_0^1 {\left( {\frac{4}{{1 + {t^2}}}j + \frac{{2t}}{{1 + {t^2}}}k} \right)} dt\)

Short answer

Therefore, the solution is \(\int_0^1 {\left( {\frac{4}{{1 + {t^2}}}j + \frac{{2t}}{{1 + {t^2}}}k} \right)} dt = 0{\bf{i}} + \pi {\bf{j}} + \ln 2{\bf{k}}\)

Step-by-step solution

Step 1: Given information

The given value is:

\(\int_0^1 {\left( {\frac{4}{{1 + {t^2}}}{\bf{j}} + \frac{{2t}}{{1 + {t^2}}}{\bf{k}}} \right)} dt\)

Step 2: Evaluate the provided integral.

Let us evaluate the given integral component-wise as shown below,\(\int_0^1 0 {\bf{i}} + \frac{4}{{1 + {t^2}}}{\bf{j}} + \frac{{2t}}{{1 + {t^2}}}{\bf{k}}dt = \int_0^1 0 dt{\bf{i}} + \int_0^1 {\frac{4}{{1 + {t^2}}}} dt{\bf{j}} + \int_0^1 {\frac{{2t}}{{1 + {t^2}}}} dt{\bf{k}}\)

\( = 0{\bf{i}} + \left( {4{{\tan }^{ - 1}}t} \right)_0^1{\bf{j}} + \left( {\ln \left| {1 + {t^2}} \right|} \right)_0^1{\bf{k}}\)

\( = 0{\bf{i}} + \pi {\bf{j}} + \ln 2{\bf{k}}\)

\(\int_0^1 {\left( {\frac{4}{{1 + {t^2}}}j + \frac{{2t}}{{1 + {t^2}}}k} \right)} dt = 0{\bf{i}} + \pi {\bf{j}} + \ln 2{\bf{k}}\)