What is the integral of 1/(sqrt(9x^2-36x+37)) ?

The integral of 1/(sqrt(9x^2-36x+37)) is 1/3 ln|3(x-2)+sqrt(9x^2-36x+37)|+C

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integral of 1/(sqrt(9x^2-36x+37))

\int \frac{1}{9 x ^{2} - 36 x + 37} d x

\frac{1}{3} ln 3 (x - 2) + 9 x^{2} - 36 x + 37 + C

Solution steps

\int \frac{1}{9 x ^{2} - 36 x + 37} d x

Complete the square

9 x^{2} - 36 x + 37 : 9 (x - 2)^{2} + 1

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{3} d v

= \int \frac{sec ( v )}{3} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int sec (v) d v

Use the common integral:

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

= \frac{1}{3} ln ∣ tan (v) + sec (v) ∣

Substitute back

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

Simplify

\frac{1}{3} ln ∣ tan (arctan (3 (x - 2))) + sec (arctan (3 (x - 2))) ∣ : \frac{1}{3} ln 3 (x - 2) + 9 x^{2} - 36 x + 37

= \frac{1}{3} ln 3 (x - 2) + 9 x^{2} - 36 x + 37

Add a constant to the solution

= \frac{1}{3} ln 3 (x - 2) + 9 x^{2} - 36 x + 37 + C

\int \frac{x ^{7} e ^{x} - 4 x ^{6}}{x ^{7}} d x

\int \frac{ln ( x )}{x ^{7}} d x

y^{^{''}} - y^{^{'}} - 2 y = - 8 t + 6 t^{2}

x \to 2 - lim (\frac{3 x + 5}{x - 2})

\int \frac{2 t}{1 + t ^{2}} d t

What is the integral of 1/(sqrt(9x^2-36x+37)) ?
The integral of 1/(sqrt(9x^2-36x+37)) is 1/3 ln|3(x-2)+sqrt(9x^2-36x+37)|+C