MHB Whitman 8.4.7 integrate by parts

[karush]

$\tiny{Whitman \ \ \ 8.4.7 }$
$\displaystyle
\int x \arctan\left({x}\right) \ dx
=\frac {{x}^{2} \arctan\left({x}\right)
+\arctan\left({x}\right)
- x} {2 } + C$
$\displaystyle uv-\int v \ du $

$\begin{align}\displaystyle
u& = x& dv&= \arctan\left({x}\right) \ dx \\
du&=dx& v& =\frac{1}{{x}^{2 }+1}
\end{align}$
So..
$$\frac{x}{{x}^{2 }+1} - \int\frac{1}{{x}^{2 }+1} \ dx$$
But this doesn't seem to be going toward the answer...

 

[MarkFL]

Your value for $v$ is incorrect...you've differentiated rather than integrated.

Try using the LIATE rule...it essentially tells you to try the inverse trig. function as your $u$ in this case.

 

[karush]

Why would I integrate u

The book examples were

u dv
du v

 

[MarkFL]

OK I see
Got distracted by the Hillary nom pom.

LIATE rule ?

Yes, this rule tells you that when faced with a choice for candidates for your $u$ in IBP, to try functions in this order:

• Logarithmic
• Inverse Trig.
• Algebraic
• Trigonometric
• Exponential

You integrand here has an algebraic function and an inverse trig. function, so try letting $u=\arctan(x)$.

Here's more on the LIATE rule:

LIATE Rule

 

[karush]

$\tiny{Whitman \ \ \ 8.4.7 }$
$\displaystyle
I=\int x \arctan\left({x}\right) \ dx
=\frac {{x}^{2} \arctan\left({x}\right)
+\arctan\left({x}\right)
- x} {2 } + C$
$\displaystyle uv-\int v \ du $

$\begin{align}\displaystyle
u& = \arctan\left({x}\right) &
dv&=x \ dx \\
du&=\frac{1}{{x}^{2 }+1}
dx& v& =\frac{{x}^{2 }}{2}
\end{align}$
So..
$$\frac{{x}^{2 }}{2} \arctan\left({x}\right)
-\frac{1}{2 } \int \frac{{x}^{2 }}{{x}^{2 }+1} \ dx$$
Then
$$I=\frac {{x}^{2} \arctan\left({x}\right)
+\arctan\left({x}\right)
- x} {2 } + C$$