Evaluate-1-3-x-2-1-x-dx-

Question Number 75081 by Rio Michael last updated on 07/Dec/19

E v a l u a t e

\int_{1}^{3} \frac{x^{2}}{1 + x} d x

Answered by mr W last updated on 07/Dec/19

\int_{1}^{3} \frac{x^{2}}{1 + x} d x

= \int_{1}^{3} \frac{x^{2} - 1 + 1}{1 + x} d x

= \int_{1}^{3} (x - 1 + \frac{1}{1 + x}) d x

= {[\frac{x^{2}}{2} - x + \ln (x + 1)]}_{1}^{3}

= \frac{3^{2} - 1^{2}}{2} - 3 + 1 + \ln (3 + 1) - \ln 2

= 2 + \ln 2

Commented by Rio Michael last updated on 07/Dec/19

t h a n k y o u s i r