How do you evaluate the integral #int arctansqrtx#?

Question

Emilia Aguilar · Accepted Answer

#intarctan(sqrtx)dx=(x+1)arctan(sqrtx)-sqrtx+C#

#I=intarctan(sqrtx)dx#

Let #t=sqrtx#. Finding #dx# in a usable form is simpler if we first write that #t^2=x#, which then implies that #2tdt=dx#. This way we can plug in #dx# into the integral straight away. These substitutions yield:

#I=intarctan(t)(2tdt)=int2tarctan(t)dt#

Now use integration by parts. Let:

#{(u=arctan(t)" "=>" "du=1/(t^2+1)dt),(dv=2tdt" "=>" "v=t^2):}#

Then:

#I=t^2arctan(t)-intt^2/(t^2+1)dt#

Rewriting the integrand as #(t^2+1-1)/(t^2+1)=(t^2+1)/(t^2+1)-1/(t^2+1)=1-1/(t^2+1)# we see that

#I=t^2arctan(t)-(intdt-int1/(t^2+1)dt)#

Both of which are common integrals:

#I=t^2arctan(t)-t+arctan(t)+C#

Returning to #x# from #t=sqrtx#:

#I=xarctan(sqrtx)-sqrtx+arctan(sqrtx)+C#

#I=(x+1)arctan(sqrtx)-sqrtx+C#

Axel Alvarez · Answer

undefined To evaluate the integral $\int \arctan(\sqrt{x}) \, dx$, we can use integration by parts. Let $ u = \arctan(\sqrt{x}) $ and $ dv = dx $. Then, $ du = \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}} \, dx $ and $ v = x $.

Applying the integration by parts formula:

$$
\int u \, dv = uv - \int v \, du
$$

we get:

$$
\begin{aligned}
\int \arctan(\sqrt{x}) \, dx &= x\arctan(\sqrt{x}) - \int \frac{x}{1 + x} \cdot \frac{1}{2\sqrt{x}} \, dx \
&= x\arctan(\sqrt{x}) - \frac{1}{2} \int \frac{\sqrt{x}}{1 + x} \, dx
\end{aligned}
$$

Now, for the remaining integral $\int \frac{\sqrt{x}}{1 + x} \, dx$, we can use a substitution method, letting $ t = \sqrt{x} $ which implies $ x = t^2 $ and $ dx = 2t \, dt $. Substituting these into the integral:

$$
\begin{aligned}
\int \frac{\sqrt{x}}{1 + x} \, dx &= \int \frac{t \cdot 2t}{1 + t^2} \, dt \
&= 2 \int \frac{t^2}{1 + t^2} \, dt
\end{aligned}
$$

This integral can be evaluated using a trigonometric substitution, letting $ t = \tan(\theta) $, so $ dt = \sec^2(\theta) \, d\theta $. Substituting these, we have:

$$
\begin{aligned}
2 \int \frac{t^2}{1 + t^2} \, dt &= 2 \int \frac{\tan^2(\theta) \cdot \sec^2(\theta)}{1 + \tan^2(\theta)} \, d\theta \
&= 2 \int \frac{\tan^2(\theta) \cdot \sec^2(\theta)}{\sec^2(\theta)} \, d\theta \
&= 2 \int \tan^2(\theta) \, d\theta \
&= 2 \int (\sec^2(\theta) - 1) \, d\theta \
&= 2 (\tan(\theta) - \theta) + C
\end{aligned}
$$

Now, we can back-substitute and integrate to get the final result:

$$
\begin{aligned}
\int \arctan(\sqrt{x}) \, dx &= x\arctan(\sqrt{x}) - \frac{1}{2} \int \frac{\sqrt{x}}{1 + x} \, dx \
&= x\arctan(\sqrt{x}) - \frac{1}{2} \left( 2(\tan(\theta) - \theta) \right) + C \
&= x\arctan(\sqrt{x}) - \tan(\theta) + \theta + C
\end{aligned}
$$

Remembering that $ x = t^2 $ and $ t = \tan(\theta) $, we can replace $ \tan(\theta) $ with $ \sqrt{x} $ and $ \theta $ with $ \arctan(\sqrt{x}) $:

$$
\int \arctan(\sqrt{x}) \, dx = x\arctan(\sqrt{x}) - \sqrt{x} + \arctan(\sqrt{x}) + C
$$