Question

How do you find the indefinite integral of `int 2*x*arctan(7*x) dx`?

Answer 1

`x^2arctan(7x)-1/7 x+1/49 arctan(7x)+C`

Explanation:

Use integration-by-parts to start doing this integral. Let `u=arctan(7x)` so that `du=1/(1+(7x)^2) * 7=7/(1+49x^2)` and let `dv=2x\ dx` so that `v=x^2`. Then

`int 2x arctan(7x)\ dx=uv-int\ v\ du=x^2arctan(7x)-int (7x^2)/(1+49x^2)`

To do this second integral, first use long division of polynomials to write `(7x^2)/(1+49x^2)=1/7-(1/7)/(1+49x^2)` and

`int (7x^2)/(1+49x^2)\ dx=1/7 x-1/7 int 1/(1+49x^2)\ dx`

On this last integral, do a substitution: `w=7x`, `dw=7dx` to write `int 1/(1+49x^2)\ dx=1/7 int 1/(1+w^2)\ dw=1/7 arctan(w)+C`.

Now you can put everything together as follows:

`int 2x arctan(7x)\ dx=x^2arctan(7x)-1/7 x+1/49 arctan(7x)+C`.

You should take the time to check this answer by differentiation (using the product rule and chain rule).