Search any question & find its solution
Question:
Answered & Verified by Expert
$\int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x=$
Options:
Solution:
1031 Upvotes
Verified Answer
The correct answer is:
$\frac{x^5}{5}+x^3+6 x+c$
$\int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x$
degree of numerator is greater than degree of denominator so,
$x^8-9 x^2+18$ is divided by $x^4-3 x^2+3$
We get quotient $x^4+3 x^2+6$ and 0 remainder, so
$\begin{aligned} & \int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x=\int\left(x^4+3 x^2+6\right) d x \\ & =\frac{x^5}{5}+\frac{3 x^3}{3}+6 x+c=\frac{x^5}{5}+x^3+6 x+c .\end{aligned}$
degree of numerator is greater than degree of denominator so,
$x^8-9 x^2+18$ is divided by $x^4-3 x^2+3$
We get quotient $x^4+3 x^2+6$ and 0 remainder, so
$\begin{aligned} & \int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x=\int\left(x^4+3 x^2+6\right) d x \\ & =\frac{x^5}{5}+\frac{3 x^3}{3}+6 x+c=\frac{x^5}{5}+x^3+6 x+c .\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.