What is the integral of (8t^3+10)/(4t^2-4t+5) ?

The integral of (8t^3+10)/(4t^2-4t+5) is t^2+2t-1/4 ln|4t^2-4t+5|-1/4 arctan(t-1/2)+C

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integral of (8t^3+10)/(4t^2-4t+5)

\int \frac{8 t ^{3} + 10}{4 t ^{2} - 4 t + 5} d t

t^{2} + 2 t - \frac{1}{4} ln 4 t^{2} - 4 t + 5 - \frac{1}{4} arctan (t - \frac{1}{2}) + C

Solution steps

\int \frac{8 t ^{3} + 10}{4 t ^{2} - 4 t + 5} d t

Long division

\frac{8 t ^{3} + 10}{4 t ^{2} - 4 t + 5} : 2 t + 2 - \frac{2 t}{4 t ^{2} - 4 t + 5}

= \int 2 t + 2 - \frac{2 t}{4 t ^{2} - 4 t + 5} 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 t d t + \int 2 d t - \int \frac{2 t}{4 t ^{2} - 4 t + 5} d t

\int 2 t d t = t^{2}

\int 2 d t = 2 t

\int \frac{2 t}{4 t ^{2} - 4 t + 5} d t = \frac{1}{4} ln 4 t^{2} - 4 t + 5 + \frac{1}{4} arctan (t - \frac{1}{2})

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

Simplify

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

Add a constant to the solution

= t^{2} + 2 t - \frac{1}{4} ln 4 t^{2} - 4 t + 5 - \frac{1}{4} arctan (t - \frac{1}{2}) + C

What is the integral of (8t^3+10)/(4t^2-4t+5) ?
The integral of (8t^3+10)/(4t^2-4t+5) is t^2+2t-1/4 ln|4t^2-4t+5|-1/4 arctan(t-1/2)+C