Another Integration Problem - Understanding Numerator

[Jason76]

\(\displaystyle \int \dfrac{x + 3}{x^{2} + 2x + 2}dx\)

\(\displaystyle \dfrac{1}{2} \int \dfrac{(x + 1) 2dx}{(x + 1)^{2} + 1} + \int \dfrac{2dx}{(x + 1)^{2} + 1}\) - :?: What's going on with the numerator? I understand the denominator (completing the square).

For instance, how can \(\displaystyle dx\) be fixed (u substitution)? Given: \(\displaystyle u = x^{2} + 2x + 2\) and \(\displaystyle du = 2x + 2\). Note: \(\displaystyle du\) and \(\displaystyle dx\) have to be fixed before splitting the fraction into two fractions.

\(\displaystyle \dfrac{1}{2} \ln(x^{2} + 2x + 2) + 2 \arctan(x + 1) + C\) - Final answer

 

[tkhunny]

Denominator

\(\displaystyle x^{2} + 2x + 2\)

\(\displaystyle \dfrac{d}{dx}x^{2} + 2x + 2 = 2x + 2\)

Numerator

\(\displaystyle x + 3 = ½(2x + 6) = ½(2x + 2) + ½(4)\)

Entire Expression

\(\displaystyle \dfrac{1}{2}\left(\dfrac{2x+2}{x^{2} + 2x + 2} + \dfrac{4}{x^{2} + 2x + 2}\right)\)

Solve the first piece via algebraic Substitution. Solve the second piece by trigonometric substitution after completing the square.

 

[Jason76]

Thanks a lot. Got this one figured out.

 

[Jason76]

Looking at this again:

\(\displaystyle \int \dfrac{x + 3}{x^{2} + 2x + 2}dx\)

Here is another "gasping at the answer" problem where \(\displaystyle du\) is greater than \(\displaystyle dx\). Another way is better. From looking at it, we see the highest index in the numerator is LESS than the highest index in the denominator, so long division is ruled out. Perhaps we could try partial fractions or trig substitution.

 

[DrPhil]

Looking at this again:

\(\displaystyle \int \dfrac{x + 3}{x^{2} + 2x + 2}dx\)

Here is another "gasping at the answer" problem where \(\displaystyle du\) is greater than \(\displaystyle dx\). Another way is better. From looking at it, we see the highest index in the numerator is LESS than the highest index in the denominator, so long division is ruled out. Perhaps we could try partial fractions or trig substitution.

Since the denominator is an irreducible quadratic, there are no further partial fractions to do. The procedure would just split you into two integrals, one with numerator x and the other with numerator 3. The trick used was to make one of the numerators equal to the derivative of the denominator, which is 2(x + 1). Thus

\(\displaystyle x + 3 = (x + 1) + 2 = \frac{1}{2}[2(x + 1)] + 2\)

The first integral gives a logarithm, and the second integral (after you complete the square in the denominator) is solved by using a trig substitution.

If you had not noticed that shortcut, you would have gotten to exactly the same place while integrating \(\displaystyle \frac{x}{x^2+2x+2}\) using the substitution \(\displaystyle u = x^2+2x+2\)

 

[Jason76]

Could it be done this way?

\(\displaystyle \int \dfrac{x + 3}{x^{2} + 2x + 2}dx\)

\(\displaystyle \int \dfrac{x }{x^{2} + 2x + 2 } + \int \dfrac{3 }{x^{2} + 2x + 2 }\)

\(\displaystyle u = x^{2} + 2x + 2 \)

\(\displaystyle du = 2x + 2\)

Now looking at the left fraction:

\(\displaystyle 2x + 2du = x dx\)

\(\displaystyle du = \dfrac{x}{2x + 2}dx\)

How would you divide those this fraction?

 

[DrPhil]

Could it be done this way?

\(\displaystyle \int \dfrac{x + 3}{x^{2} + 2x + 2}dx\)

\(\displaystyle \int \dfrac{x }{x^{2} + 2x + 2 } + \int \dfrac{3 }{x^{2} + 2x + 2 }\)......don't forget the "dx" in each integration

\(\displaystyle u = x^{2} + 2x + 2 \)

\(\displaystyle du = 2x + 2\)......du/dx = 2x + 2

\(\displaystyle 2x + 2du = x dx\)......X

\(\displaystyle du = \dfrac{x}{2x + 2}dx\)......X

Maybe try some long division to get the constant of integration for the left fraction?

You can rearrange du/dx to find (x dx) for substitution in the integrand:
\(\displaystyle x\ dx = (1/2) du - dx \)
The new numerator has two terms, one a function of u and the other a function of x. That numerator of dx will combine with the second fraction from above.

\(\displaystyle \displaystyle \int \dfrac{(1/2)du}{u} - \int \dfrac{dx}{x^{2} + 2x + 2 } + \int \dfrac{3\ dx }{x^{2} + 2x + 2 }\)

\(\displaystyle \displaystyle \dfrac{1}{2}\int \dfrac{du}{u} + 2\int \dfrac{dx}{x^{2} + 2x + 2 }\)

As promised, this is exactly like the shortcut introduced by tkhunny:

\(\displaystyle x + 3 = \frac{1}{2}(2x + 2) + 2\)