What is the integral of 1/(4t^2-4t+5) ?

The integral of 1/(4t^2-4t+5) is 1/4 arctan(t-1/2)+C

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integral of 1/(4t^2-4t+5)

\int \frac{1}{4 t ^{2} - 4 t + 5} d t

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

Solution steps

\int \frac{1}{4 t ^{2} - 4 t + 5} d t

Complete the square

4 t^{2} - 4 t + 5 : 4 (t - \frac{1}{2})^{2} + 4

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

Apply u-substitution

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

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

\frac{1}{4 u ^{2} + 4} = \frac{1}{4} \frac{1}{u ^{2} + 1}

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{4} arctan (u)

Substitute back

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (- 4 y^{3} sin (4 x))

\frac{d}{d x} (sin (x) - csc (x))

\int (2 x^{2} - 3)^{2} d x

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

cos (x) + i sin (x)

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