Solve the integral

Question

Integral of √ x^2 − 9/ x^2 dx

note: the numerator x^2-9 is all under the square root

Answer 1

We use sabstitution: x = 3secu; dx = 3secutanudu∫

∫(x² - 9)^1/2/x²dx = ∫3tanu·3tanusecudu/9sec²u = ∫tan²ucosudu = ∫sin²u/cosudu = ∫(secu - cosu)du =

ln(secu + tanuI - sinu + C; secu = x/3; cosu = 3/x; sinu = √1 - 9/x² = √x² - 9/xtanu = √x² - 9/3∫

∫(x² - 9)^1/2/x²dx = ln[Ix/3 + √x² - 9/3I - √x² - 9/x + C = lnIx + √x² - 9I - √x² - 9/x + C.

We include - ln3 in C

Answer 2

I = ∫ √(x²-9) /x² dx

Use integration by parts

u = √(x²-9) du = x/√(x²-9)

dv = 1/x² v = -1/x

I = -√(x²-9) /x - ∫ -1/√(x²-9) dx

Focus on ∫ 1/√(x²-9) dx

let u = x/3 du = dx/3 dx = 3du x = 3u

∫1/√(u²+1) du

Let u = secθ du = secθtanθdθ

∫secθtanθ/√(tan²θ+1) dθ = ∫secθ dθ which is a known integral

= ln|tanθ+secθ| = ln|√(u²-1) + u|

= ln|√(x²/9-1) + x/3|

Putting both parts together

∫ √(x²-9) /x² dx = ln|√(x²/9-1) + x/3| - √(x²-9) /x +C