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

Question

Ezra Adams · Accepted Answer

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

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).

Luke Alvarez · Answer

undefined To find the indefinite integral of $ \int 2x \cdot \arctan(7x) \, dx $, you can use integration by parts.

The formula for integration by parts is:
$$ \int u \, dv = uv - \int v \, du $$

In this case, let:
$$ u = \arctan(7x) $$
$$ dv = 2x \, dx $$

Then, calculate $ du $ and $ v $:
$$ du = \frac{1}{1 + (7x)^2} \cdot 7 \, dx = \frac{7}{1 + 49x^2} \, dx $$
$$ v = \int 2x \, dx = x^2 $$

Now, apply the integration by parts formula:
$$ \int 2x \cdot \arctan(7x) \, dx = uv - \int v \, du $$
$$ = x^2 \arctan(7x) - \int x^2 \cdot \frac{7}{1 + 49x^2} \, dx $$

The integral on the right side, $ \int x^2 \cdot \frac{7}{1 + 49x^2} \, dx $, can be solved using a trigonometric substitution:

Let $ t = 7x $, then $ dt = 7 \, dx $, and $ x^2 = \frac{t^2}{49} $.

Substituting:
$$ \int x^2 \cdot \frac{7}{1 + 49x^2} \, dx = \int \frac{t^2/49 \cdot 7}{1 + t^2} \cdot \frac{1}{7} \, dt $$
$$ = \frac{1}{49} \int \frac{t^2}{1 + t^2} \, dt $$

Now, use another trigonometric substitution $ t = \tan(\theta) $, $ dt = \sec^2(\theta) \, d\theta $:
$$ = \frac{1}{49} \int \frac{\tan^2(\theta)}{1 + \tan^2(\theta)} \cdot \sec^2(\theta) \, d\theta $$
$$ = \frac{1}{49} \int \sin^2(\theta) \, d\theta $$

This integral can be solved using a trigonometric identity:
$$ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} $$

Substituting back:
$$ \frac{1}{49} \int \frac{1 - \cos(2\theta)}{2} \, d\theta $$
$$ = \frac{1}{49} \left(\frac{\theta}{2} - \frac{\sin(2\theta)}{4}\right) + C $$

Finally, substitute back $ \theta = \arctan(7x) $ and $ t = 7x $ to get the final answer:
$$ \int 2x \cdot \arctan(7x) \, dx = x^2 \arctan(7x) - \frac{1}{98} \left(\arctan(7x) - \frac{7x}{1 + 49x^2}\right) + C $$
$$ = x^2 \arctan(7x) - \frac{1}{98} \arctan(7x) + \frac{1}{98} \cdot \frac{7x}{1 + 49x^2} + C $$

Where $ C $ is the constant of integration.