Use partial fractions to evaluate the definite integral. ∫01x2−xx2+x+1dx

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Use partial fractions to evaluate the definite integral.   ∫01x2−xx2+x+1dx

Dawn Neal · Accepted Answer

Step 1: Consider the provided definite integral,  ∫01x2−xx2+x+1dx  Now, use the partial fraction to reduce the integral and after that integrate it.  ∫01x2−xx2+x+1dx=∫01(−2x−1x2+x+1+1)dx  =∫01−2x−1x2+x+1dx+∫011dx  =[−ln⁡(x2+x+1)]01+[x]01  =[−ln⁡(1+1+1)−ln⁡0]+1  =−ln⁡3+1  Step 2:   ∫01x2−xx2+x+1dx≈−ln⁡3+1  Hence, the result is verified.  The result of given definite integral is true.

stomachdm · Accepted Answer

∫x2−xx2+x+1dx  The degree of the numerator P(x) is greater than or equal to the degree of the denominator Q(x), so we divide the polynomials.  x2−xx2+x+1=1+−2x−1x2+x+1  Integrating the integer part, we get:  ∫1⋅dx=x  Integrating further, we get:  ∫−2x−1x2+x+1dx=−ln⁡(x2+x+1)  Answer:  x−ln⁡(x2+x+1)+C  Calculate the definite integral:  ∫01x2−xx2+x+1dx=(x−ln⁡(x2+x+1))∣01  F(1)=1−ln⁡(3)  F(0)=0  I=1−ln⁡(3)−(0)=1−ln⁡(3)

Use partial fractions to evaluate the definite integral. \int_{0}^{1}\frac{x^{2}-x}{x^{2}+x+1}dx

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