What is the integral of ((8t^3+13))/((t+2)(4t^2+1)) ?

The integral of ((8t^3+13))/((t+2)(4t^2+1)) is 2t-3ln|t+2|-1/2 ln|4t^2+1|+3arctan(2t)+C

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integral of ((8t^3+13))/((t+2)(4t^2+1))

\int \frac{( 8 t ^{3} + 13 )}{( t + 2 ) ( 4 t ^{2} + 1 )} d t

2 t - 3 ln ∣ t + 2 ∣ - \frac{1}{2} ln 4 t^{2} + 1 + 3 arctan (2 t) + C

Solution steps

\int \frac{8 t ^{3} + 13}{( t + 2 ) ( 4 t ^{2} + 1 )} d t

Take the partial fraction of

\frac{8 t ^{3} + 13}{( t + 2 ) ( 4 t ^{2} + 1 )} : 2 - \frac{3}{t + 2} + \frac{- 4 t + 6}{4 t ^{2} + 1}

= \int 2 - \frac{3}{t + 2} + \frac{- 4 t + 6}{4 t ^{2} + 1} d t

Apply the Sum Rule:

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

= \int 2 d t - \int \frac{3}{t + 2} d t + \int \frac{- 4 t + 6}{4 t ^{2} + 1} d t

\int 2 d t = 2 t

\int \frac{3}{t + 2} d t = 3 ln ∣ t + 2 ∣

\int \frac{- 4 t + 6}{4 t ^{2} + 1} d t = - \frac{1}{2} ln 4 t^{2} + 1 + 3 arctan (2 t)

= 2 t - 3 ln ∣ t + 2 ∣ - \frac{1}{2} ln 4 t^{2} + 1 + 3 arctan (2 t)

Add a constant to the solution

= 2 t - 3 ln ∣ t + 2 ∣ - \frac{1}{2} ln 4 t^{2} + 1 + 3 arctan (2 t) + C

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What is the integral of ((8t^3+13))/((t+2)(4t^2+1)) ?
The integral of ((8t^3+13))/((t+2)(4t^2+1)) is 2t-3ln|t+2|-1/2 ln|4t^2+1|+3arctan(2t)+C