Calculus

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int(dx/(x^2+9)) u = x/3

Use the indicated substitution (above) to evaluate the integral. Confirm answer by differentiation.

Okay, so I found that du/dx is 1/3. dx is 3du. Just by looking at the integral I can tell this is some form of tan^-1(x), if we change it to int(1/(x^2+9) * dx). Thing is, I don't really know where to put in u to make this work. Help please???

Thank you!

  • Calculus -

    u=x/3
    du = (1/3)dx
    dx = 3du
    x²=9u²
    Substitute in the integral to get:
    ∫ dx/(x²+9)
    =∫ 3du/(9u²+9)
    =∫ (1/3)du/(x²+1)
    Can you take it from here, using your tan-1 suggestion?

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