Determine whether the following substitution is correct. The integral of 1/(9+x^2) dx. Let x = 3/t. Then....?

Question

Then, dx/dt = -3/t^2.

The integral of 1/(9+x^2) dx = the integral of (-3/t^2)/(9+(3/t)^2) dt = (-1/3)*the integral of 1/(1+t^2) dt = (-1/3)*arctan(t) + C = (-1/3)*arctan(3/x) + C.

I know we can just derive the integral using the formula for the derivative of tangent inverse. But will this substitution work? Is it correct?

Thanks!

Answer 1

Try x = 3t
dx = 3 dt

I = ∫ 1 / ( 9 + 9 t ² ) 3 dt
I = (1/3) ∫ 1 / (1 + t ²) dt
I = (1/3) tan ^(-1) t + C
I = (1/3) tan^(-1) (x / 3) + C

Answer 2

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