Q.Evaluate : $\int \dfrac{1}{x^4} \, dx$

Expand the square, then integrate: $\dfrac{x^5}{5}+\dfrac{8}{3}x^3+16x+c$.

First expand: $(x^2+4)^2=x^4+8x^2+16$. Then integrate each term: …

By the power rule of integration, $\int x^5\,dx=\dfrac{x^6}{6}+C$.

Step 1 — the standard power rule of integration is

$\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C,\quad n\neq -1$

Step 2 — with $n=5$:

$\int x^{5}\,dx=\dfrac{x^{6}}{6}+C$

…

The integral is $x^3 + x + C$.

Apply the power rule of integration $\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1}$ to each term:

$\int 3x^2\,dx = 3\cdot\frac{x^3}{3} = x^3, \qquad \int 1\,dx = x.$

Therefore …

Correct option: (a) $x^2$.

Using $\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1}$,

$\int 2x\,dx = 2\cdot\frac{x^{2}}{2} = x^2\;(+\,C).$

…