evaluate the integral

Question

∫1/(x^2+9)^2 dx

evaluate the integral in terms of arctan

Answer 1

The presence of (x² + 9) suggests trying a trigonometric substitution, specifically:

x = 3 tan θ

dx = 3 sec² θ dθ

x² + 9 = 9 tan² θ + 9 = 9(tan² θ + 1) = 9 sec² θ

Hence the integrand is transformed into:

3 sec² θ dθ / (9 sec² θ)² dθ = dθ/(27sec² θ) = (1/27)cos² θ dθ

You should already know how to compute:

∫ cos² θ dθ = (1/2) [θ + sin(2θ)/2] = (1/2) [θ + sin(θ)cos(θ)]

To transform back in terms of x...

From x = 3 tan θ, we get θ = arctan(x/3). To get the sine and cosine in terms of x, draw a right triangle and label one of the acute angles as θ, the opposite side as x, and the adjacent side as 3. The hypotenuse is thus √(x² + 9) by the Pythagorean Theorem. This gives:

sinθ = x / √(x² + 9)

cosθ = 3 / √(x² + 9)

...and you can put all the pieces together to finish. Don't forget the (1/27) in front.