What is the integral of 8/(t(t^2+2)^2) ?

The integral of 8/(t(t^2+2)^2) is 2ln(t)-ln|t^2+2|+2/(t^2+2)+C

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integral of 8/(t(t^2+2)^2)

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

2 ln (t) - ln t^{2} + 2 + \frac{2}{t ^{2} + 2} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 8 \cdot \int \frac{1}{2 u ( u + 2 ) ^{2}} d u

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{u ( u + 2 ) ^{2}} d u

Take the partial fraction of

\frac{1}{u ( u + 2 ) ^{2}} : \frac{1}{4 u} - \frac{1}{4 ( u + 2 )} - \frac{1}{2 ( u + 2 ) ^{2}}

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{4 u} - \frac{1}{4 ( u + 2 )} - \frac{1}{2 ( u + 2 ) ^{2}} 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

= 8 \cdot \frac{1}{2} (\int \frac{1}{4 u} d u - \int \frac{1}{4 ( u + 2 )} d u - \int \frac{1}{2 ( u + 2 ) ^{2}} d u)

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

\int \frac{1}{4 ( u + 2 )} d u = \frac{1}{4} ln ∣ u + 2 ∣

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

= 8 \cdot \frac{1}{2} (\frac{1}{4} ln ∣ u ∣ - \frac{1}{4} ln ∣ u + 2 ∣ - (- \frac{1}{2 ( u + 2 )}))

Substitute back

u = t^{2}

= 8 \cdot \frac{1}{2} (\frac{1}{4} ln t^{2} - \frac{1}{4} ln t^{2} + 2 - (- \frac{1}{2 ( t ^{2} + 2 )}))

Simplify

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

= 2 ln (t) - ln t^{2} + 2 + \frac{2}{t ^{2} + 2}

Add a constant to the solution

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

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What is the integral of 8/(t(t^2+2)^2) ?
The integral of 8/(t(t^2+2)^2) is 2ln(t)-ln|t^2+2|+2/(t^2+2)+C