Answer
$\tan ti+\frac{1}{8}(t^2+1)^4j+(\dfrac{1}{3}t^3 \ln t-\dfrac{1}{9}t^3)k+C$
Work Step by Step
As we are given that $\int (\sec^2 ti+t(t^2+1)^3j+t^2 \ln t k)dt$
Suppose $I=\int (\sec^2 ti+t(t^2+1)^3j+t^2 \ln t k)dt$
$I=\int \sec^2 tdti+\int t(t^2+1)^3dtj+\int t^2 \ln t dtk$
$I=\tan ti+ Pj+Qk$ ... (a)
Here, $P=\int t(t^2+1)^3dt$ and $Q=\int t^2 \ln t dt$
i) Consider $P=\int t(t^2+1)^3dt$
Put $p=t^2+1 $ gives $dp=2tdt$
Thus, $P=\frac{1}{2}\int p^3dp=\frac{1}{8}p^4=\frac{1}{8}(t^2+1)^4$
ii) $Q=\int t^2 \ln t dt=\ln t (1/3t^3)-\int (1/3t^3) \dfrac{1}{t} dt=\dfrac{1}{3}t^3 \ln t-\dfrac{1}{9}t^3$
From equation (a):
$I=\tan ti+\frac{1}{8}(t^2+1)^4j+(\dfrac{1}{3}t^3 \ln t-\dfrac{1}{9}t^3)k+C$; (C is the constant of integration)
