Factor the denominator and split the rational function by partial fractions.
1 x 3 + x = 1 x ( x 2 + 1 ) = A x + B x + C x 2 + 1 . \frac{1}{x^{3}+x}=\frac{1}{x(x^{2}+1)}=\frac{A}{x}+\frac{Bx+C}{x^{2}+1}. x 3 + x 1 = x ( x 2 + 1 ) 1 = x A + x 2 + 1 B x + C .
Clearing denominators: 1 = A ( x 2 + 1 ) + ( B x + C ) x = ( A + B ) x 2 + C x + A 1=A(x^{2}+1)+(Bx+C)x=(A+B)x^{2}+Cx+A 1 = A ( x 2 + 1 ) + ( B x + C ) x = ( A + B ) x 2 + C x + A A = 1 , C = 0 , B = − 1 A=1,\ C=0,\ B=-1 A = 1 , C = 0 , B = − 1
1 x ( x 2 + 1 ) = 1 x − x x 2 + 1 . \frac{1}{x(x^{2}+1)}=\frac{1}{x}-\frac{x}{x^{2}+1}. x ( x 2 + 1 ) 1 = x 1 − x 2 + 1 x .
Antiderivative:
∫ ( 1 x − x x 2 + 1 ) d x = ln ∣ x ∣ − 1 2 ln ( x 2 + 1 ) + C . \int\left(\frac{1}{x}-\frac{x}{x^{2}+1}\right)dx=\ln|x|-\tfrac{1}{2}\ln(x^{2}+1)+C. ∫ ( x 1 − x 2 + 1 x ) d x = ln ∣ x ∣ − 2 1 ln ( x 2 + 1 ) + C .
Evaluate from 1 1 1 2 2 2
[ ln x − 1 2 ln ( x 2 + 1 ) ] 1 2 = ( ln 2 − 1 2 ln 5 ) − ( 0 − 1 2 ln 2 ) = 3 2 ln 2 − 1 2 ln 5. \left[\ln x-\tfrac{1}{2}\ln(x^{2}+1)\right]_{1}^{2}=\left(\ln 2-\tfrac{1}{2}\ln 5\right)-\left(0-\tfrac{1}{2}\ln 2\right)=\tfrac{3}{2}\ln 2-\tfrac{1}{2}\ln 5. [ ln x − 2 1 ln ( x 2 + 1 ) ] 1 2 = ( ln 2 − 2 1 ln 5 ) − ( 0 − 2 1 ln 2 ) = 2 3 ln 2 − 2 1 ln 5.
Combine into one logarithm:
1 2 ( 3 ln 2 − ln 5 ) = 1 2 ln 8 5 = ln 8 5 . \tfrac{1}{2}\left(3\ln 2-\ln 5\right)=\tfrac{1}{2}\ln\frac{8}{5}=\ln\sqrt{\frac{8}{5}}. 2 1 ( 3 ln 2 − ln 5 ) = 2 1 ln 5 8 = ln 5 8 .
This is option 2. (Option 3 would arise from ∫ 1 2 1 x 2 + 1 d x \int_{1}^{2}\frac{1}{x^{2}+1}dx ∫ 1 2 x 2 + 1 1 d x
Exam trick: When a denominator is x ( x 2 + 1 ) x(x^{2}+1) x ( x 2 + 1 ) x x 2 + 1 \frac{x}{x^{2}+1} x 2 + 1 x 1 2 ln ( x 2 + 1 ) \tfrac{1}{2}\ln(x^{2}+1) 2 1 ln ( x 2 + 1 ) 1 2 \tfrac12 2 1 ln ⋅ \ln\sqrt{\cdot} ln ⋅