What is the integral of x/(sqrt(x^2-14x+45)) ?

The integral of x/(sqrt(x^2-14x+45)) is sqrt(x^2-14x+45)+7ln(1/2 |sqrt(x^2-14x+45)+x-7|)+C

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integral of x/(sqrt(x^2-14x+45))

\int \frac{x}{x ^{2} - 14 x + 45} d x

x^{2} - 14 x + 45 + 7 ln (\frac{1}{2} x^{2} - 14 x + 45 + x - 7) + C

Solution steps

\int \frac{x}{x ^{2} - 14 x + 45} d x

Complete the square

x^{2} - 14 x + 45 : (x - 7)^{2} - 4

= \int \frac{x}{( x - 7 ) ^{2} - 4} d x

Apply u-substitution

= \int \frac{u + 7}{u ^{2} - 4} d u

Expand

\frac{u + 7}{u ^{2} - 4} : \frac{u}{u ^{2} - 4} + \frac{7}{u ^{2} - 4}

= \int \frac{u}{u ^{2} - 4} + \frac{7}{u ^{2} - 4} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{u}{u ^{2} - 4} d u + \int \frac{7}{u ^{2} - 4} d u

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

\int \frac{7}{u ^{2} - 4} d u = 7 ln (\frac{1}{2} u^{2} - 4 + u)

= u^{2} - 4 + 7 ln (\frac{1}{2} u^{2} - 4 + u)

Substitute back

u = x - 7

= (x - 7)^{2} - 4 + 7 ln (\frac{1}{2} (x - 7)^{2} - 4 + x - 7)

Simplify

(x - 7)^{2} - 4 + 7 ln (\frac{1}{2} (x - 7)^{2} - 4 + x - 7) : x^{2} - 14 x + 45 + 7 ln (\frac{1}{2} x^{2} - 14 x + 45 + x - 7)

= x^{2} - 14 x + 45 + 7 ln (\frac{1}{2} x^{2} - 14 x + 45 + x - 7)

Add a constant to the solution

= x^{2} - 14 x + 45 + 7 ln (\frac{1}{2} x^{2} - 14 x + 45 + x - 7) + C

What is the integral of x/(sqrt(x^2-14x+45)) ?
The integral of x/(sqrt(x^2-14x+45)) is sqrt(x^2-14x+45)+7ln(1/2 |sqrt(x^2-14x+45)+x-7|)+C