Question

Find \(\rm \int {1\over x^2}dx\) = ?
1. \(\rm {1\over x} +C\)
2. \(\rm {-1\over x} +C\)
3. \(\rm {1\over 2x} +C\)
4. \(\rm {-1\over 2x} +C\)

Answer 1

Correct Answer - Option 2 : \(\rm {-1\over x} +C\)

Concept:

Integral property:

• ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
• \(\rm∫ {1\over x} dx = \ln x\) + C
• ∫ ex dx = ex+ C
• ∫ ax dx = (ax/ln a) + C ; a > 0,  a ≠ 1
• ∫ sin x dx = - cos x + C
• ∫ cos x dx = sin x + C

Calculation:

I = \(\rm ∫ {1\over x^2}dx\)

I = \(\rm ∫ x^{-2}dx\)

I = \(\rm {x^{(-2+1)}\over(-2+1)} +C\)

I = \(\rm {x^{-1}\over-1} +C\)

I = \(\boldsymbol{\rm -{1\over x}+C}\)