Evaluate each of the following integral: `int_0^3 1/(x^2+9)dx`

Explanation

Substituting `x=3 tan theta` `Rightarrow dx=3 sec ^{2} theta d theta` (By differentiating both sides)

Also, when `x=0, theta=0` and `x=3, theta=frac{pi}{4}`

We get


`I=int_{0}^{frac{pi}{4}} frac{1}{9+9 tan ^{2} theta} 3 sec ^{2} theta d theta`


Since `sec ^{2} theta=1+tan ^{2} theta`

We get `I=int_{0}^{pi / 4} frac{d theta}{3}`
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