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?
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!
1 answer
1 answer