$\int {\frac{{\left( {{x^2} + 2} \right)d}}{{x + 1}}} x$
$\int {\frac{{\left( {{x^2} + 2} \right)d}}{{x + 1}}} x$
Official Solution
VVidaara Team
✓ Verified solution
NCERT & Exemplar
}
Let $I = \int {\frac{{{x^2} + 2}}{{x + 1}}} dx$
$= \int {\left( {x - 1 + \frac{3}{{x + 1}}} \right)} dx$
$= \int {(x - 1)} dx + 3\int {\frac{1}{{x + 1}}} dx$
$= \frac{{{x^2}}}{2} - x + 3\log |(x + 1)| + C$
Community Answers (0)
Log in to post your own answer or join the discussion.
No comments yet — start the discussion.