What is the integral of (4t)/((t^2+3)^2) ?

The integral of (4t)/((t^2+3)^2) is -2/(t^2+3)+C

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integral of (4t)/((t^2+3)^2)

\int \frac{4 t}{( t ^{2} + 3 ) ^{2}} d t

- \frac{2}{t ^{2} + 3} + C

Solution steps

\int \frac{4 t}{( t ^{2} + 3 ) ^{2}} d t

Take the constant out:

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

= 4 \cdot \int \frac{t}{( t ^{2} + 3 ) ^{2}} d t

Apply u-substitution

= 4 \cdot \int \frac{1}{2 u ^{2}} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

= 4 \cdot \frac{1}{2} (- \frac{1}{u})

Substitute back

u = t^{2} + 3

= 4 \cdot \frac{1}{2} (- \frac{1}{t ^{2} + 3})

Simplify

4 \cdot \frac{1}{2} (- \frac{1}{t ^{2} + 3}) : - \frac{2}{t ^{2} + 3}

= - \frac{2}{t ^{2} + 3}

Add a constant to the solution

= - \frac{2}{t ^{2} + 3} + C

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