What is the integral of (t^4+9)/(t^4+9t^2) ?

The integral of (t^4+9)/(t^4+9t^2) is t-1/t-10/3 arctan(t/3)+C

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integral of (t^4+9)/(t^4+9t^2)

\int \frac{t ^{4} + 9}{t ^{4} + 9 t ^{2}} d t

t - \frac{1}{t} - \frac{10}{3} arctan (\frac{t}{3}) + C

Solution steps

\int \frac{t ^{4} + 9}{t ^{4} + 9 t ^{2}} d t

Take the partial fraction of

\frac{t ^{4} + 9}{t ^{4} + 9 t ^{2}} : 1 + \frac{1}{t ^{2}} - \frac{10}{t ^{2} + 9}

= \int 1 + \frac{1}{t ^{2}} - \frac{10}{t ^{2} + 9} 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{1}{t ^{2}} d t - \int \frac{10}{t ^{2} + 9} d t

\int 1 d t = t

\int \frac{1}{t ^{2}} d t = - \frac{1}{t}

\int \frac{10}{t ^{2} + 9} d t = \frac{10}{3} arctan (\frac{t}{3})

= t - \frac{1}{t} - \frac{10}{3} arctan (\frac{t}{3})

Add a constant to the solution

= t - \frac{1}{t} - \frac{10}{3} arctan (\frac{t}{3}) + C

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What is the integral of (t^4+9)/(t^4+9t^2) ?
The integral of (t^4+9)/(t^4+9t^2) is t-1/t-10/3 arctan(t/3)+C