Question
$\int {\frac{{\left( {{x^2} + 2} \right)d}}{{x + 1}}} x$
$\int {\frac{{\left( {{x^2} + 2} \right)d}}{{x + 1}}} x$
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Let $I = \int {\frac{{{x^2} + 2}}{{x + 1}}} dx$
$= \int {\left( {x - 1 + \frac{3}{{x + 1}}} \right)} dx$
$= \int {(x - 1)} dx + 3\int {\frac{1}{{x + 1}}} dx$
$= \frac{{{x^2}}}{2} - x + 3\log |(x + 1)| + C$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Integrals. Curated by Sachin Sharma. Free for all students.