What is the integral of (t^2(t^2+6))/((t^2+2)^2) ?

The integral of (t^2(t^2+6))/((t^2+2)^2) is t-1/(sqrt(2))sin(2arctan(t/(sqrt(2))))+C

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integral of (t^2(t^2+6))/((t^2+2)^2)

\int \frac{t ^{2} ( t ^{2} + 6 )}{( t ^{2} + 2 ) ^{2}} d t

t - \frac{1}{2} sin (2 arctan (\frac{t}{2})) + C

Solution steps

\int \frac{t ^{2} ( t ^{2} + 6 )}{( t ^{2} + 2 ) ^{2}} d t

Take the partial fraction of

\frac{t ^{2} ( t ^{2} + 6 )}{( t ^{2} + 2 ) ^{2}} : 1 + \frac{2}{t ^{2} + 2} - \frac{8}{( t ^{2} + 2 ) ^{2}}

= \int 1 + \frac{2}{t ^{2} + 2} - \frac{8}{( t ^{2} + 2 ) ^{2}} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d t + \int \frac{2}{t ^{2} + 2} d t - \int \frac{8}{( t ^{2} + 2 ) ^{2}} d t

\int 1 d t = t

\int \frac{2}{t ^{2} + 2} d t = 2 arctan (\frac{t}{2})

\int \frac{8}{( t ^{2} + 2 ) ^{2}} d t = 2 (arctan (\frac{t}{2}) + \frac{1}{2} sin (2 arctan (\frac{t}{2})))

= t + 2 arctan (\frac{t}{2}) - 2 (arctan (\frac{t}{2}) + \frac{1}{2} sin (2 arctan (\frac{t}{2})))

Simplify

t + 2 arctan (\frac{t}{2}) - 2 (arctan (\frac{t}{2}) + \frac{1}{2} sin (2 arctan (\frac{t}{2}))) : t - \frac{1}{2} sin (2 arctan (\frac{t}{2}))

= t - \frac{1}{2} sin (2 arctan (\frac{t}{2}))

Add a constant to the solution

= t - \frac{1}{2} sin (2 arctan (\frac{t}{2})) + C

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What is the integral of (t^2(t^2+6))/((t^2+2)^2) ?
The integral of (t^2(t^2+6))/((t^2+2)^2) is t-1/(sqrt(2))sin(2arctan(t/(sqrt(2))))+C