What is the integral of 6/(sqrt(9-(x+3)^2)) ?

The integral of 6/(sqrt(9-(x+3)^2)) is 6arcsin(1/3 x+1)+C

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integral of 6/(sqrt(9-(x+3)^2))

\int \frac{6}{9 - ( x + 3 ) ^{2}} d x

6 arcsin (\frac{1}{3} x + 1) + C

Solution steps

\int \frac{6}{9 - ( x + 3 ) ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{9 - ( x + 3 ) ^{2}} d x

Apply u-substitution

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

= 6 \cdot \int \frac{1}{9 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 1 d v

= 6 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 6 \cdot 1 \cdot v

Substitute back

= 6 \cdot 1 \cdot arcsin (\frac{1}{3} (x + 3))

Simplify

6 \cdot 1 \cdot arcsin (\frac{1}{3} (x + 3)) : 6 arcsin (\frac{1}{3} x + 1)

= 6 arcsin (\frac{1}{3} x + 1)

Add a constant to the solution

= 6 arcsin (\frac{1}{3} x + 1) + C

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What is the integral of 6/(sqrt(9-(x+3)^2)) ?
The integral of 6/(sqrt(9-(x+3)^2)) is 6arcsin(1/3 x+1)+C