Correct Answer - Option 1 :
\(\rm -{1\over2x^2}-\log x+{1\over2}\log(x^2+1)\) + C
Concept:
Integral property:
- ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
-
\(\rm∫ {1\over x} dx = \ln x\) + C
Calculation:
I = \(\rm\int{dx\over x^5 +x^3}\)
I = \(\rm\int{dx\over x^3(x^2 +1)}\)
By partial fraction
I = \(\rm \int {x\over x^2+1}-{1\over x}+{1\over x^3}\) dx
I = \(\rm \int {x\over x^2+1}dx-\int {dx\over x}+\int{dx\over x^3}\)
Let x2 + 1 = t ⇒ 2x dx = dt
I = \(\rm \int {dt\over2t}-\ln x+\int x^{-3}dx + C\)
I = \(\rm {1\over2}\ln t-\ln x+{x^{-2}\over-2} + C\)
Putting the value of t
I = \(\rm {1\over2}\ln (x^2+1)-\ln x-{1\over 2x^2} + C\)