What is the integral of (t^5)/((t^3+1)^2) ?

The integral of (t^5)/((t^3+1)^2) is 1/3 (ln|t^3+1|+1/(t^3+1))+C

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integral of (t^5)/((t^3+1)^2)

\int \frac{t ^{5}}{( t ^{3} + 1 ) ^{2}} d t

\frac{1}{3} (ln t^{3} + 1 + \frac{1}{t ^{3} + 1}) + C

Solution steps

\int \frac{t ^{5}}{( t ^{3} + 1 ) ^{2}} d t

Apply u-substitution

: \int \frac{u ^{\frac{1}{5}}}{5 ( u ^{\frac{3}{5}} + 1 ) ^{2}} d u

= \int \frac{u ^{\frac{1}{5}}}{5 ( u ^{\frac{3}{5}} + 1 ) ^{2}} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{u ^{\frac{1}{5}}}{( u ^{\frac{3}{5}} + 1 ) ^{2}} d u

Apply u-substitution

: \int \frac{5 ( v - 1 )}{3 v ^{2}} d v

= \frac{1}{5} \cdot \int \frac{5 ( v - 1 )}{3 v ^{2}} d v

Take the constant out:

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

= \frac{1}{5} \cdot \frac{5}{3} \cdot \int \frac{v - 1}{v ^{2}} d v

Expand

\frac{v - 1}{v ^{2}} : \frac{1}{v} - \frac{1}{v ^{2}}

= \frac{1}{5} \cdot \frac{5}{3} \cdot \int \frac{1}{v} - \frac{1}{v ^{2}} d v

Apply the Sum Rule:

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

= \frac{1}{5} \cdot \frac{5}{3} (\int \frac{1}{v} d v - \int \frac{1}{v ^{2}} d v)

\int \frac{1}{v} d v = ln ∣ v ∣

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

= \frac{1}{5} \cdot \frac{5}{3} (ln ∣ v ∣ - (- \frac{1}{v}))

Substitute back

= \frac{1}{5} \cdot \frac{5}{3} (ln (t^{5})^{\frac{3}{5}} + 1 - (- \frac{1}{( t ^{5} ) ^{\frac{3}{5}} + 1}))

Simplify

\frac{1}{5} \cdot \frac{5}{3} (ln (t^{5})^{\frac{3}{5}} + 1 - (- \frac{1}{( t ^{5} ) ^{\frac{3}{5}} + 1})) : \frac{1}{3} (ln t^{3} + 1 + \frac{1}{t ^{3} + 1})

= \frac{1}{3} (ln t^{3} + 1 + \frac{1}{t ^{3} + 1})

Add a constant to the solution

= \frac{1}{3} (ln t^{3} + 1 + \frac{1}{t ^{3} + 1}) + C

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What is the integral of (t^5)/((t^3+1)^2) ?
The integral of (t^5)/((t^3+1)^2) is 1/3 (ln|t^3+1|+1/(t^3+1))+C