\(\displaystyle \int \dfrac{du}{\sqrt{u^{2} - a^{2}}} = \cosh^{-1} \dfrac{u}{a} + C\)
Spotting: We notice a square root in the denominator and the numbers resemble the above formula. In this case, not so easily recognizable, unless the a term can be manipulated so there would be a negative sign (-) in front of it.
\(\displaystyle \int \dfrac{dx}{\sqrt{x^{2} + 6x}}\) Complete the square
\(\displaystyle \int \dfrac{dx}{\sqrt{x^{2} + 6x + 9 - 9}}\)
\(\displaystyle \int \dfrac{dx}{\sqrt{(x + 3)^{2} - 9}}\)
\(\displaystyle u = x + 3 \)
\(\displaystyle du = (1)dx \)
\(\displaystyle (1)du = dx \) (since 1 no need for constant of integration)
\(\displaystyle a = 3\)
\(\displaystyle \int \dfrac{dx}{\sqrt{(x + 3)^{2} - (3)^{2}}} = cosh^{-1} \dfrac{x + 3}{3} + C\) Final answer
Look right?
Spotting: We notice a square root in the denominator and the numbers resemble the above formula. In this case, not so easily recognizable, unless the a term can be manipulated so there would be a negative sign (-) in front of it.
\(\displaystyle \int \dfrac{dx}{\sqrt{x^{2} + 6x}}\) Complete the square
\(\displaystyle \int \dfrac{dx}{\sqrt{x^{2} + 6x + 9 - 9}}\)
\(\displaystyle \int \dfrac{dx}{\sqrt{(x + 3)^{2} - 9}}\)
\(\displaystyle u = x + 3 \)
\(\displaystyle du = (1)dx \)
\(\displaystyle (1)du = dx \) (since 1 no need for constant of integration)
\(\displaystyle a = 3\)
\(\displaystyle \int \dfrac{dx}{\sqrt{(x + 3)^{2} - (3)^{2}}} = cosh^{-1} \dfrac{x + 3}{3} + C\) Final answer
Look right?
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Forgot the "completing square thing". So 6x is divided by 2 and then squared?