What is the integral of (24t+6t^3)/(8t^2+t^4) ?

The integral of (24t+6t^3)/(8t^2+t^4) is 3ln(t)+3/2 ln|t^2+8|+C

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integral of (24t+6t^3)/(8t^2+t^4)

\int \frac{24 t + 6 t ^{3}}{8 t ^{2} + t ^{4}} d t

3 ln (t) + \frac{3}{2} ln t^{2} + 8 + C

Solution steps

\int \frac{24 t + 6 t ^{3}}{8 t ^{2} + t ^{4}} d t

Expand

\frac{24 t + 6 t ^{3}}{8 t ^{2} + t ^{4}} : \frac{24 t}{8 t ^{2} + t ^{4}} + \frac{6 t ^{3}}{8 t ^{2} + t ^{4}}

Apply the Sum Rule:

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

= \int \frac{24 t}{8 t ^{2} + t ^{4}} d t + \int \frac{6 t ^{3}}{8 t ^{2} + t ^{4}} d t

\int \frac{24 t}{8 t ^{2} + t ^{4}} d t = 24 (\frac{1}{8} ln (t) - \frac{1}{16} ln t^{2} + 8)

\int \frac{6 t ^{3}}{8 t ^{2} + t ^{4}} d t = 3 ln t^{2} + 8

= 24 (\frac{1}{8} ln (t) - \frac{1}{16} ln t^{2} + 8) + 3 ln t^{2} + 8

Simplify

24 (\frac{1}{8} ln (t) - \frac{1}{16} ln t^{2} + 8) + 3 ln t^{2} + 8 : 3 ln (t) + \frac{3}{2} ln t^{2} + 8

= 3 ln (t) + \frac{3}{2} ln t^{2} + 8

Add a constant to the solution

= 3 ln (t) + \frac{3}{2} ln t^{2} + 8 + C

e^{- 2 t - 3}

\frac{d}{d x} (\frac{x}{π})

\frac{d}{d x} (arccos (y - 2 x))

\int sec^{2} (θ) tan^{7} (θ) d θ

x \to 2 lim (\frac{- 1}{( x + 2 )})

What is the integral of (24t+6t^3)/(8t^2+t^4) ?
The integral of (24t+6t^3)/(8t^2+t^4) is 3ln(t)+3/2 ln|t^2+8|+C