Dividing numerator and denominator by $x^{15}$ we get,
$ = \int {\frac{{\frac{2}{{{x^3}}} + \frac{5}{{{x^6}}}}}{{{{\left( {1 + \frac{1}{{{x^2}}} + \frac{1}{{{x^5}}}} \right)}^3}}}} dx$
Put $\left(1+\frac{1}{x^{2}}+\frac{1}{x^{5}}\right)=t$
$=\int \frac{-d t}{t^{3}}$
$=\frac{-t^{-3+1}}{-3+1}+C=\frac{1}{2} \times \frac{1}{t^{2}}+C$
$=\frac{1}{2} \frac{1}{\left(1+\frac{1}{x^{2}}+\frac{1}{x^{5}}\right)^{2}}+C$
$=\frac{1}{2} \frac{x^{10}}{\left(x^{5}+x^{3}+1\right)^{2}}+C$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$f(x) = \left\{ {\begin{array}{*{20}{c}}
{{x^2} + 2mx - 1\,,}&{x \leq 0}\\
{mx - 1\,\,\,\,\,\,\,\,\,\,\,\,\,,}&{x > 0}
\end{array}} \right.$
જો $f (x)$ એક-એક વિધેય હોય તો $'m'$ ની કિમતોનો ગણ મેળવો.
