1.

Find the value of \(\rm \int{3\over x^2+4x-5}\) dx1.  \(\rm {1\over2}\ln \left|{x-1\over x+5} \right|\) + C2.  \(\rm \ln \left|{x-1\over x+5} \right|\) + C3.  \(\rm {1\over2}\) [ln |x-1| + ln |x+5|] + C4.  [ln |x-1| + ln |x+5| + C

Answer» Correct Answer - Option 1 :  \(\rm {1\over2}\ln \left|{x-1\over x+5} \right|\) + 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 = \(\rm \int{3\over x^2+4x-5}\) dx

⇒ I = \(\rm 3\int{1\over x^2+4x-5}\) dx

⇒ I = \(\rm 3\int{1\over (x-1)(x+5)}\) dx

⇒ I = \(\rm 3\int{(x+5)-(x-1)\over6(x-1)(x+5)}\) dx

⇒ I = \(\rm {3\over6}\int{(x+5)\over (x-1)(x+5)}-{(x-1)\over (x-1)(x+5)}\) dx

⇒ I = \(\rm {1\over2}\int{1\over (x-1)}-{1\over (x+5)}\) dx

⇒ I = \(\rm {1\over2}\) [ln |x-1| - ln |x+5|] + C

⇒ I = \(\boldsymbol{\rm {1\over2}\ln \left|{x-1\over x+5} \right|}\) + C


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