JEE Main 12 Jan 2019 Evening Question 18
Question: The integral $ \int _{{}}^{{}}{\frac{3x^{13}+2x^{11}}{{{(2x^{4}+3x^{2}+1)}^{4}}}}dx $ is equal to (where C is a constant of integration) [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ \frac{x^{4}}{6{{(2x^{4}+3x^{2}+1)}^{3}}}+C $
B) $ \frac{x^{4}}{{{(2x^{4}+3x^{2}+1)}^{3}}}+C $
C) $ \frac{x^{12}}{6{{(2x^{4}+3x^{2}+1)}^{3}}}+C $
D) $ \frac{x^{12}}{{{(2x^{4}+3x^{2}+1)}^{3}}}+C $
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Answer:
Correct Answer: C
Solution:
-
Let $ I=\int _{{}}^{{}}{\frac{3x^{13}+2x^{11}}{{{(2x^{4}+3x^{2}+1)}^{4}}}}dx $ $ =\int _{{}}^{{}}{\frac{\frac{3}{x^{3}}+\frac{2}{x^{5}}}{{{( 2+\frac{3}{x^{2}}+\frac{1}{x^{4}} )}^{4}}}}dx $
Now, put $ 2+\frac{3}{x^{2}}+\frac{1}{x^{4}}=z $
$ \therefore $ $ ( \frac{-6}{x^{3}}-\frac{4}{x^{5}} )dx=dz $
$ \Rightarrow $ $ ( \frac{3}{x^{3}}+\frac{2}{x^{5}} )dx=\frac{-dz}{2} $
$ \therefore $ $ I=\frac{-1}{2}\int _{{}}^{{}}{\frac{dz}{z^{4}}=\frac{-1}{2}}( \frac{{z^{-3}}}{-3} )+C $ $ =\frac{1}{6z^{3}}+C=\frac{1}{6{{( 2+\frac{3}{x^{2}}+\frac{1}{x^{4}} )}^{3}}}+C $ $ =\frac{x^{12}}{6{{(2x^{4}+3x^{2}+1)}^{3}}}+C $
