Mathematics

# Solve : $\int \dfrac{t^4 - 3 t^2 + 2}{t^2 (1 + t^2)} , dt$

##### SOLUTION
$I=\displaystyle \int \dfrac {t^4 -3t^2 +2}{t^2 (1+t^2)}dt$
$=\displaystyle \int \dfrac {t^4 +t^2 4t^2+2}{t^2 (1+t^2)}dt$
$\Rightarrow \ \displaystyle \int dt-\displaystyle \int \dfrac {4t^2}{t^2 (1+t^2)}dt +\displaystyle \int \dfrac {2}{t^2 (1+t^2)}dt$
$I\Rightarrow \ t-\displaystyle \int \dfrac {4}{(1+t^2)}dt+\displaystyle \int 2 \left[\dfrac {1}{t^2}-\dfrac {1}{1+t^2}\right] dt$
$I\Rightarrow \ t-4\tan^{-1} (t)+2 \displaystyle \int \dfrac {1}{t^2}dt+\displaystyle \int \dfrac {2}{(1+t^2)}dt$
$I=t-4\tan^{-1}t+2\left(\dfrac {t^{-2+1}}{-2+1}\right)-2\tan^{-1} (t)+C$
$\boxed {I\Rightarrow \ t-\dfrac {2}{t}-6\tan^{-1} (t)+C}$
$\therefore \ \boxed {\displaystyle \int \dfrac {t^4-3t^2+2}{t^2 (1+t^2)}dt=t-\dfrac {2}{t}-6\tan^{-1} (t)+C}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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