What is the integral of 4/(sqrt(x^2-4x+13)) ?

The integral of 4/(sqrt(x^2-4x+13)) is 4ln(1/3 |x-2+sqrt(x^2-4x+13)|)+C

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integral of 4/(sqrt(x^2-4x+13))

\int \frac{4}{x ^{2} - 4 x + 13} d x

4 ln (\frac{1}{3} x - 2 + x^{2} - 4 x + 13) + C

Solution steps

\int \frac{4}{x ^{2} - 4 x + 13} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} - 4 x + 13} d x

Complete the square

x^{2} - 4 x + 13 : (x - 2)^{2} + 9

= 4 \cdot \int \frac{1}{( x - 2 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 9} d u

= 4 \cdot \int \frac{1}{u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= 4 \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= 4 ln ∣ tan (v) + sec (v) ∣

Substitute back

= 4 ln tan (arctan (\frac{1}{3} (x - 2))) + sec (arctan (\frac{1}{3} (x - 2)))

Simplify

4 ln tan (arctan (\frac{1}{3} (x - 2))) + sec (arctan (\frac{1}{3} (x - 2))) : 4 ln (\frac{1}{3} x - 2 + x^{2} - 4 x + 13)

= 4 ln (\frac{1}{3} x - 2 + x^{2} - 4 x + 13)

Add a constant to the solution

= 4 ln (\frac{1}{3} x - 2 + x^{2} - 4 x + 13) + C

\int_{0}^{5} \frac{1}{x - 3} d x

θ \to 0 lim (\frac{θ}{sin ^{2} ( θ )})

x \to 2 lim (x^{2} - 7 x + 12)

\frac{1}{( s ^{3} + s ^{2} + s )}

x \to 0 lim (\frac{9 sin ( 1 )}{1})

What is the integral of 4/(sqrt(x^2-4x+13)) ?
The integral of 4/(sqrt(x^2-4x+13)) is 4ln(1/3 |x-2+sqrt(x^2-4x+13)|)+C