Answer
$\mathop \smallint \limits_0^1 \left( {t{{\rm{e}}^{ - {t^2}}}{\bf{i}} + t\ln \left( {{t^2} + 1} \right){\bf{j}}} \right){\rm{d}}t = \left( { - \frac{1}{2}\left( {{{\rm{e}}^{ - 1}} - 1} \right)} \right){\bf{i}} + \left( {\ln 2 - \frac{1}{2}} \right){\bf{j}}$
Work Step by Step
Evaluate $\mathop \smallint \limits_0^1 \left( {t{{\rm{e}}^{ - {t^2}}}{\bf{i}} + t\ln \left( {{t^2} + 1} \right){\bf{j}}} \right){\rm{d}}t$.
Since vector-valued integrals obey the same linearity rules as scalar-valued integrals, so
$\mathop \smallint \limits_0^1 \left( {t{{\rm{e}}^{ - {t^2}}}{\bf{i}} + t\ln \left( {{t^2} + 1} \right){\bf{j}}} \right){\rm{d}}t = \left( {\mathop \smallint \limits_0^1 t{{\rm{e}}^{ - {t^2}}}{\rm{d}}t} \right){\bf{i}}$
${\ \ \ \ }$ $ + \left( {\mathop \smallint \limits_0^1 t\ln \left( {{t^2} + 1} \right){\rm{d}}t} \right){\bf{j}}$
Evaluate $\mathop \smallint \limits_0^1 t{{\rm{e}}^{ - {t^2}}}{\rm{d}}t$
Write $v = {t^2}$. So, $dv = 2tdt$. The integral becomes
$\mathop \smallint \limits_0^1 t{{\rm{e}}^{ - {t^2}}}{\rm{d}}t = \frac{1}{2}\cdot\mathop \smallint \limits_0^1 {{\rm{e}}^{ - v}}{\rm{d}}v = - \frac{1}{2}{{\rm{e}}^{ - v}}|_0^1$
$\mathop \smallint \limits_0^1 t{{\rm{e}}^{ - {t^2}}}{\rm{d}}t = - \frac{1}{2}\left( {{{\rm{e}}^{ - 1}} - 1} \right)$
Evaluate $\mathop \smallint \limits_0^1 t\ln \left( {{t^2} + 1} \right){\rm{d}}t$
Write $v = {t^2} + 1$. So, $dv = 2tdt$. The integral becomes
$\mathop \smallint \limits_0^1 t\ln \left( {{t^2} + 1} \right){\rm{d}}t = \frac{1}{2}\mathop \smallint \limits_1^2 \ln v{\rm{d}}v$
Recall from integration by parts formula:
$\smallint u{\rm{d}}v = uv - \smallint v{\rm{d}}u$
Write $u = \ln v$. So, $du = \frac{1}{v}dv$. The integral becomes
$\frac{1}{2}\mathop \smallint \limits_1^2 \ln v{\rm{d}}v = \frac{1}{2}\left[ {\left( {v\ln v} \right)|_1^2 - \mathop \smallint \limits_1^2 v\cdot\frac{1}{v}{\rm{d}}v} \right]$
$\frac{1}{2}\mathop \smallint \limits_1^2 \ln v{\rm{d}}v = \frac{1}{2}\left[ {\left( {v\ln v} \right)|_1^2 - v|_1^2} \right]$
$\frac{1}{2}\mathop \smallint \limits_1^2 \ln v{\rm{d}}v = \frac{1}{2}\left[ {2\ln 2 - \ln 1 - 1} \right] = \frac{1}{2}\left( {2\ln 2 - 1} \right)$
$\frac{1}{2}\mathop \smallint \limits_1^2 \ln v{\rm{d}}v = \ln 2 - \frac{1}{2}$
From the results above we obtain
$\mathop \smallint \limits_0^1 \left( {t{{\rm{e}}^{ - {t^2}}}{\bf{i}} + t\ln \left( {{t^2} + 1} \right){\bf{j}}} \right){\rm{d}}t = \left( { - \frac{1}{2}\left( {{{\rm{e}}^{ - 1}} - 1} \right)} \right){\bf{i}} + \left( {\ln 2 - \frac{1}{2}} \right){\bf{j}}$
