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

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

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

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

\frac{1}{8} (ln 4 t^{2} - 4 t + 3 - 2 arctan (2 (t - \frac{1}{2}))) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{2 u - 1}{2 ( 4 u ^{2} + 2 )} d u

= \int \frac{2 u - 1}{2 ( 4 u ^{2} + 2 )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{2 u - 1}{4 u ^{2} + 2} d u

Expand

\frac{2 u - 1}{4 u ^{2} + 2} : \frac{2 u}{4 u ^{2} + 2} - \frac{1}{4 u ^{2} + 2}

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}{2} (\int \frac{2 u}{4 u ^{2} + 2} d u - \int \frac{1}{4 u ^{2} + 2} d u)

\int \frac{2 u}{4 u ^{2} + 2} d u = \frac{1}{4} ln 4 u^{2} + 2

\int \frac{1}{4 u ^{2} + 2} d u = \frac{1}{2 2} arctan (2 u)

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

Substitute back

u = t - \frac{1}{2}

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

Simplify

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

= \frac{1}{8} (ln 4 t^{2} - 4 t + 3 - 2 arctan (2 (t - \frac{1}{2})))

Add a constant to the solution

= \frac{1}{8} (ln 4 t^{2} - 4 t + 3 - 2 arctan (2 (t - \frac{1}{2}))) + C

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