Given: $I=\int x^5 \sec^2 (7x^6+9) dx$
Let $7x^6+9=t$
Differentiating both the sides
$42 x^5 dx=dt$
$\Rightarrow\:x^5\:dx=\large\frac{1}{42}$$dt$
Substituting the values of $t$ and $dx$ in I we get
$\Rightarrow\:I=\int \large\frac{1}{42} $$\sec^2t +c $
$\Rightarrow\:I=\large\frac{1}{42} $$\tan t+c $ (Since $\int sec^2t dt=tant$)
Substituting the value of $t$ we get
$\Rightarrow\:I=\large\frac{1}{42}$$ \tan (7x^6 +9)+c $
Hence (a) is the correct answer.