The integral ₀¹ ⁻¹(1+x+x^2)dx is equal to:

Step-by-step solution

I = ₀¹ ⁻¹(1+x+x^2)dx I = ₀¹ ⁻¹ ( 1 1+x+x^2 )dx I = ₀¹ ⁻¹ ( (x+1)-x 1+(x+1)x )dx I = ₀¹( ⁻¹(x+1)- ⁻¹x)dx I = ₀¹ ⁻¹(x+1)dx - ₀¹ ⁻¹x dx Substituting x+1 = t in the first integral: I = ₁² ⁻¹x dx - ₀¹ ⁻¹x dx Using integration by parts, ⁻¹x dx = x ⁻¹x - 1 2 _e(1+x^2) I = [x ⁻¹x - 1 2 _e(1+x^2) ]₁² - [x ⁻¹x - 1 2 _e(1+x^2) ]₀¹ I = (2 ⁻¹2 - 1 2 _e 5 - ( 4 - 1 2 _e 2 ) ) - ( 4 - 1 2 _e 2 - 0 ) I = 2 ⁻¹2 - 1 2 _e 5 - 2 + _e 2 I = 2 ⁻¹2 - 1 2 _e 5 + 1 2 _e 4 - 2 I = 2 ⁻¹2 - 1 2 _e ( 5 4 ) - 2 Answer: 2 ⁻¹2- 1 2 _e ( 5 4 )- 2