Question

evaluate integral [(sqrt(1+x^2))/x^2]

Answer 1

\[\int\limits \frac{ \sqrt{1+x^{2}} }{ x^{2} } dx\]

Answer 2

did you try \(\large x = \tan \theta \) ?

Answer 3

yes, x = tanθ so the expression in the sqrt is 1+tan^2θ which is equal to sec^2θ.
dx = sec^2θ dθ
so the whole thing would look like integral (secθ sec^2θ / tan^2θ) dθ

Answer 4

\[\int\limits \frac{ \sqrt{\sec^2\theta} }{ \tan^2 \theta } \sec^2\theta d \theta\]

Answer 5

looks good, simplify
maybe change everything into sin/cos and cancel whatever u can

Answer 6

\[\int\limits \sec \theta \csc \theta d \theta \]

Answer 7

im not sure how i would integral that function.

Answer 8

check again you should get :
\[\large \int \sec \theta \csc^2 \theta d\theta\]

Answer 9

do you have to do something with the d(theta)?

Answer 10

write csc^2 as 1+cot^2

Answer 11

\[\large \begin{align}\int \sec \theta \csc^2 \theta d\theta &= \int \sec\theta (1+\cot^2\theta) d\theta\\~\\&=\int \sec\theta d\theta + \int \dfrac{\cos \theta }{\sin^2\theta}d\theta \end{align}\]

Answer 12

integrating them should be straightforward

Answer 13

The way i am learning this is by making the integral function easy to integrate. it will still contain (theta) in the function but you can connect theta with x from when we let x = tan(theta).

Answer 14

yeah evaluate the integral first