What is the integral of 1/(9-(3x+5)^2) ?

The integral of 1/(9-(3x+5)^2) is 1/18 (ln|(3x+5)/3+1|-ln|(3x+5)/3-1|)+C

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integral of 1/(9-(3x+5)^2)

\int \frac{1}{9 - ( 3 x + 5 ) ^{2}} d x

\frac{1}{18} (ln \frac{3 x + 5}{3} + 1 - ln \frac{3 x + 5}{3} - 1) + C

Solution steps

\int \frac{1}{9 - ( 3 x + 5 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{3 ( 9 - u ^{2} )} d u

= \int \frac{1}{3 ( 9 - u ^{2} )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{9 - u ^{2}} d u

Apply Integral Substitution

= \frac{1}{3} \cdot \int \frac{1}{3 ( - 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}{3} \cdot \frac{1}{3} \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{3} \cdot \frac{1}{3} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= \frac{1}{3} \cdot \frac{1}{3} (\frac{ln \frac{3 x + 5}{3} + 1}{2} - \frac{ln \frac{3 x + 5}{3} - 1}{2})

Simplify

\frac{1}{3} \cdot \frac{1}{3} (\frac{ln \frac{3 x + 5}{3} + 1}{2} - \frac{ln \frac{3 x + 5}{3} - 1}{2}) : \frac{1}{18} (ln \frac{3 x + 5}{3} + 1 - ln \frac{3 x + 5}{3} - 1)

= \frac{1}{18} (ln \frac{3 x + 5}{3} + 1 - ln \frac{3 x + 5}{3} - 1)

Add a constant to the solution

= \frac{1}{18} (ln \frac{3 x + 5}{3} + 1 - ln \frac{3 x + 5}{3} - 1) + C

y^{^{'''}} - 4 y^{^{''}} + y^{^{'}} + 6 y = 36

x \to - 1 lim (x^{2} + x)

\int_{2}^{5} \frac{1}{x - 2} d x

\int e^{4 x} cos (x) d x

f (x) = - 2 x^{2} - 1, at x = 1

What is the integral of 1/(9-(3x+5)^2) ?
The integral of 1/(9-(3x+5)^2) is 1/18 (ln|(3x+5)/3+1|-ln|(3x+5)/3-1|)+C