Question

integration of dx/x^2 ...Plz help me ​

Answer 1

Answer:

$\int {\frac{1}{ x^{2} } \, dx$ = $\frac{-1}{3x^{3} }$ +c

Step-by-step explanation:

We need to calculate $\int {\frac{1}{ x^{2} } \, dx$

We know that $\frac{1}{x^{2} }$ can be written as x⁻²

Therefore, $\int {\frac{1}{ x^{2} } \, dx$ = $\int { x^{-2} } \, dx$

We know that $\int { x^{n} } \, dx$ = $\frac{x^{n+1} }{n+1}$

Therefore, we will solve $\int { x^{-2} } \, dx$ with the above formula,

We will take n = -2

=> $\int {\frac{1}{ x^{2} } \, dx$

= $\int { x^{-2} } \, dx$

= $\frac{x^{-2-1} }{-2-1}$ + c

= $\frac{x^{-3} }{-3}$ + c

= -$\frac{1}{3}$ * $\frac{1}{x^{3} }$ + c

= $\frac{-1}{3x^{3} }$ +c

Therefore, $\int {\frac{1}{ x^{2} } \, dx$ = $\frac{-1}{3x^{3} }$ +c