206.8.7.58 Int 1/(x^2-6x+34) dx complete the square

[karush]

$\text{206.8.7.58}$
$\text{given and evaluation}$
$$\displaystyle
I_{58}=\int \frac{dx}{{x}^{2}-6x+34}
=\dfrac{\arctan\left(\frac{x-3}{5}\right)}{5}
+ C$$
$\text{complete the square} $
$${x}^{2}-6x+34 = \left(x-3\right)^2 + 5^2 = {u}^{2}+{a}^{2} \\
u=x-3 \\ a=5$$
$\text{standard integral} $
$$\displaystyle
I_{17}=\int \frac{du}{{a}^{2}+{u}^{2}}
=\frac{1}{a}
\arctan{\frac{u}{a}}
+C$$
$\text{back substitute } u=x-3 \ \ a=5 $

$$\displaystyle
I_{58}=\frac{1}{5}
\arctan{\frac{x-3}{5}}
+C $$
My question on this is when is standard intregral used
this could of gone on with a trig subst.
Also why isn't $\ln\left({u^2+a^2}\right)+C$
correct

 

[MarkFL]

That standard integral was derived using a trig. substitution.

What is $$\frac{d}{du}\left(\ln\left(u^2+a^2\right)+C\right)$$?

 

[karush]

Didn't understand the $\ln$. question?

 

[MarkFL]

Didn't understand the $\ln$. question?

Given that:

$$\frac{d}{du}\left(\int f(u)\,du\right)=f(u)$$

Then, if we posit:

$$\int \frac{1}{u^2+a^2}\,du=\ln\left(u^2+a^2\right)+C$$

We should be able to verify by checking to see if:

$$\frac{d}{du}\left(\ln\left(u^2+a^2\right)+C\right)=\frac{1}{u^2+a^2}$$

If not, then we know that anti-derivative is wrong. :)

 

[HallsofIvy]

$\text{206.8.7.58}$
$\text{given and evaluation}$
$$\displaystyle
I_{58}=\int \frac{dx}{{x}^{2}-6x+34}
=\dfrac{\arctan\left(\frac{x-3}{5}\right)}{5}
+ C$$
$\text{complete the square} $
$${x}^{2}-6x+34 = \left(x+3\right)^2 + 5^2 = {u}^{2}+{a}^{2} $$

You have a mistake in this first line- it should be $(x- 3)^2+ 5^2$, not x+ 3.

$$u=x+3 \\ a=5$$
$\text{standard integral} $
$$\displaystyle
I_{17}=\int \frac{du}{{a}^{2}+{u}^{2}}
=\frac{1}{a}
\arctan{\frac{u}{a}}
+C$$
$\text{back substitute } u=x+3 \ \ a=5 $

$$\displaystyle
I_{58}=\frac{1}{5}
\arctan{\frac{x+3}{5}}
+C $$
My question on this is when is standard intregral used
this could of gone on with a trig subst.
Also why isn't $\ln\left({u^2+a^2}\right)+C$
correct

 

[karush]

$\text{206.8.7.58}$
$\text{given and evaluation}$
$$\displaystyle
I_{58}=\int \frac{dx}{{x}^{2}-6x+34}
=\dfrac{\arctan\left(\frac{x-3}{5}\right)}{5}
+ C$$
$\text{complete the square} $
$${x}^{2}-6x+34 = \left(x-3\right)^2 + 5^2 = {u}^{2}+{a}^{2} \\
u=x-3 \\ a=5$$
$\text{standard integral} $
$$\displaystyle
I_{17}=\int \frac{du}{{a}^{2}+{u}^{2}}
=\frac{1}{a}
\arctan{\frac{u}{a}}
+C$$
$\text{back substitute } u=x-3 \ \ a=5 $

$$\displaystyle
I_{58}=\frac{1}{5}
\arctan{\frac{x-3}{5}}
+C $$

$\text{fixed } u=x-3 $

 

[karush]

$\text{206.8.7.58}$
$\text{given and evaluation}$
$$\displaystyle
I_{58}=\int \frac{dx}{{x}^{2}-6x+34}
=\dfrac{\arctan\left(\frac{x-3}{5}\right)}{5}
+ C$$
$\text{complete the square} $
$${x}^{2}-6x+34 = \left(x-3\right)^2 + 25$$
$\text{u substitution} $
$$u=5\tan\left({\theta}\right)
\therefore du=5\sec^2\left(\theta\right) \, d\theta
\therefore \theta =\arctan\left[\frac{u}{5}\right]$$
$\text{then..}$
$$I_{58}=.
\int\frac{5\sec^2\theta}{25\sec^2\theta+25 } \, d\theta
=\frac{1}{5}\int 1 \,d\theta=\frac{1}{5}\theta$$

$\text{back substitute }
u=x-3 \ \ \theta = \arctan\left[\frac{u}{5}\right]$

$$\displaystyle
I_{58}=\frac{1}{5}
\arctan{\frac{x-3}{5}}
+C $$