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

The integral of (t^2)/(t^2+4) is -2arctan(t/2)+t+C

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

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

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

Solution steps

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

\frac{t ^{2}}{t ^{2} + 4} = - \frac{4}{t ^{2} + 4} + 1

= \int - \frac{4}{t ^{2} + 4} + 1 d t

Apply Integral Substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int - \frac{1}{u ^{2} + 1} + 1 d u

Apply the Sum Rule:

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

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

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int 1 d u = u

= 2 (- arctan (u) + u)

Substitute back

u = \frac{t}{2}

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

Simplify

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

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

Add a constant to the solution

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

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