What is the integral of (t^2-6t+13)^{-1} ?

The integral of (t^2-6t+13)^{-1} is 1/2 arctan((t-3)/2)+C

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integral of (t^2-6t+13)^{-1}

integral

(t^{2} - 6 t + 13)^{- 1}

\frac{1}{2} arctan (\frac{t - 3}{2}) + C

Solution steps

\int (t^{2} - 6 t + 13)^{- 1} d t

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \int \frac{1}{t ^{2} - 6 t + 13} d t

Complete the square

t^{2} - 6 t + 13 : (t - 3)^{2} + 4

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{2} arctan (v)

Substitute back

= \frac{1}{2} arctan (\frac{t - 3}{2})

Add a constant to the solution

= \frac{1}{2} arctan (\frac{t - 3}{2}) + C

\frac{( 1 + x ) ^{3}}{x}

(ln (x))^{4}

(\frac{4 + x}{2})^{3}

\frac{u}{( 1 + u ^{2} )}

\frac{2 tan ( x )}{( 1 + 2 tan ^{2} ( x ) )}

What is the integral of (t^2-6t+13)^{-1} ?
The integral of (t^2-6t+13)^{-1} is 1/2 arctan((t-3)/2)+C