What is the integral of (2t)/((t^2+2)^5) ?

The integral of (2t)/((t^2+2)^5) is (t^2)/((t^2+2)^5)-5/(4(t^2+2)^4)+2/((t^2+2)^5)+C

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integral of (2t)/((t^2+2)^5)

\int \frac{2 t}{( t ^{2} + 2 ) ^{5}} d t

\frac{t ^{2}}{( t ^{2} + 2 ) ^{5}} - \frac{5}{4 ( t ^{2} + 2 ) ^{4}} + \frac{2}{( t ^{2} + 2 ) ^{5}} + C

Solution steps

\int \frac{2 t}{( t ^{2} + 2 ) ^{5}} d t

Take the constant out:

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

= 2 \cdot \int \frac{t}{( t ^{2} + 2 ) ^{5}} d t

Apply u-substitution

: \int \frac{1}{2 ( u + 2 ) ^{5}} d u

= 2 \cdot \int \frac{1}{2 ( u + 2 ) ^{5}} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{( u + 2 ) ^{5}} d u

Apply Integration By Parts

: \frac{u}{( u + 2 ) ^{5}} - \int - \frac{5 u}{( u + 2 ) ^{6}} d u

= 2 \cdot \frac{1}{2} (\frac{u}{( u + 2 ) ^{5}} - \int - \frac{5 u}{( u + 2 ) ^{6}} d u)

\int - \frac{5 u}{( u + 2 ) ^{6}} d u = \frac{5}{4 ( u + 2 ) ^{4}} - \frac{2}{( u + 2 ) ^{5}}

= 2 \cdot \frac{1}{2} (\frac{u}{( u + 2 ) ^{5}} - (\frac{5}{4 ( u + 2 ) ^{4}} - \frac{2}{( u + 2 ) ^{5}}))

Substitute back

u = t^{2}

= 2 \cdot \frac{1}{2} (\frac{t ^{2}}{( t ^{2} + 2 ) ^{5}} - (\frac{5}{4 ( t ^{2} + 2 ) ^{4}} - \frac{2}{( t ^{2} + 2 ) ^{5}}))

Simplify

2 \cdot \frac{1}{2} (\frac{t ^{2}}{( t ^{2} + 2 ) ^{5}} - (\frac{5}{4 ( t ^{2} + 2 ) ^{4}} - \frac{2}{( t ^{2} + 2 ) ^{5}})) : \frac{t ^{2}}{( t ^{2} + 2 ) ^{5}} - \frac{5}{4 ( t ^{2} + 2 ) ^{4}} + \frac{2}{( t ^{2} + 2 ) ^{5}}

= \frac{t ^{2}}{( t ^{2} + 2 ) ^{5}} - \frac{5}{4 ( t ^{2} + 2 ) ^{4}} + \frac{2}{( t ^{2} + 2 ) ^{5}}

Add a constant to the solution

= \frac{t ^{2}}{( t ^{2} + 2 ) ^{5}} - \frac{5}{4 ( t ^{2} + 2 ) ^{4}} + \frac{2}{( t ^{2} + 2 ) ^{5}} + C

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What is the integral of (2t)/((t^2+2)^5) ?
The integral of (2t)/((t^2+2)^5) is (t^2)/((t^2+2)^5)-5/(4(t^2+2)^4)+2/((t^2+2)^5)+C