What is the integral of 1/(sqrt(9x^2+36x+35)) ?

The integral of 1/(sqrt(9x^2+36x+35)) is 1/3 ln|sqrt(9x^2+36x+35)+3(x+2)|+C

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integral of 1/(sqrt(9x^2+36x+35))

\int \frac{1}{9 x ^{2} + 36 x + 35} d x

\frac{1}{3} ln 9 x^{2} + 36 x + 35 + 3 (x + 2) + C

Solution steps

\int \frac{1}{9 x ^{2} + 36 x + 35} d x

Complete the square

9 x^{2} + 36 x + 35 : 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 (arcsec (3 (x + 2))) + sec (arcsec (3 (x + 2))) ∣

Simplify

\frac{1}{3} ln ∣ tan (arcsec (3 (x + 2))) + sec (arcsec (3 (x + 2))) ∣ : \frac{1}{3} ln 9 x^{2} + 36 x + 35 + 3 (x + 2)

= \frac{1}{3} ln 9 x^{2} + 36 x + 35 + 3 (x + 2)

Add a constant to the solution

= \frac{1}{3} ln 9 x^{2} + 36 x + 35 + 3 (x + 2) + C

y = \frac{8 x}{x ^{2} - 1}, (3, 3)

y^{^{''}} + 625 y = sec (25 x)

e^{5 x}

\int e^{- 1 + x} d x

\frac{\partial}{\partial r} (e^{r} sin (s t))

What is the integral of 1/(sqrt(9x^2+36x+35)) ?
The integral of 1/(sqrt(9x^2+36x+35)) is 1/3 ln|sqrt(9x^2+36x+35)+3(x+2)|+C