Evaluate the following integrals

gashan · Sep 24, 2010, 11:07 PM

The s like sign e(raised to the power of 2x)+x(raised to the power of 2) cos3x dx. And the other question is
The s sign 2x+1 divided by the square root of (x(raised to the power of 2)+3x+6)

Unknown008 · Sep 25, 2010, 01:11 AM

1. $\int \frac{2x+5}{x^2+3x+6} dx$

This is close to the form:

$\int \frac{f'(x)}{f(x)}dx = ln|f(x)|$

But you have f'(x) here = 2x + 3. So, break the fraction into two parts first.

2. Try a long division first as the fraction isn't a simple fraction.

3. $\int e^{2x} + x^2cos(3x) dx$

Separate into two integrals and use by parts for the x^2cos(3x) one.

4. $\int \frac{2x+1}{\sqrt{x^2+3x+6}}dx$

Use integration by parts for this one.

gashan · Sep 25, 2010, 01:19 AM

Quote:

Originally Posted by Unknown008

1. $\int \frac{2x+5}{x^2+3x+6} dx$

This is close to the form:

$\int \frac{f'(x)}{f(x)}dx = ln|f(x)|$

But you have f'(x) here = 2x + 3. So, break the fraction into two parts first.

2. Try a long division first as the fraction isn't a simple fraction.

3. $\int e^{2x} + x^2cos(3x) dx$

Separate into two integrals and use by parts for the x^2cos(3x) one.

4. $\int \frac{2x+1}{\sqrt{x^2+3x+6}}dx$

Use integration by parts for this one.

My best friend asked me to check on a solution for this question. I think he missed out a semester and is trying to teach himself from scratch. Would it be possible to give me the entire solution? Looking forward to your response

galactus · Sep 25, 2010, 04:31 AM

For the last one:

$\int\frac{2x-1}{\sqrt{x^{2}+3x+6}}dx$

Break it up into:

$\int\frac{2x+3}{\sqrt{x^{2}+3x+6}}dx-\int\frac{4}{\sqrt{x^{2}+3x+6}}dx$

For the first half, let
$u=x^{2}+3x+6, \;\ du=(2x+3)dx$

See how?. It whittles down to

$\int\frac{1}{\sqrt{u}}du$

Now, continue with the second half.

For that part, complete the square on the quadratic in the radical and then make an appropriate substitution.

The first problem can be tackled by completing the square as well.

gashan · Sep 25, 2010, 05:43 AM

Quote:

Originally Posted by galactus

For the last one:

$\int\frac{2x-1}{\sqrt{x^{2}+3x+6}}dx$

Break it up into:

$\int\frac{2x+3}{\sqrt{x^{2}+3x+6}}dx-\int\frac{4}{\sqrt{x^{2}+3x+6}}dx$

For the first half, let
$u=x^{2}+3x+6, \;\ du=(2x+3)dx$

See how?. It whittles down to

$\int\frac{1}{\sqrt{u}}du$

Now, continue with the second half.

For that part, complete the square on the quadratic in the radical and then make an appropriate substitution.

The first problem can be tackled by completing the square as well.

for the second part, would it be equal to 1/4U+C?

galactus · Sep 25, 2010, 06:05 AM

No, it is a bit more involved than that.

We start with:

$4\int\frac{1}{\sqrt{x^{2}+3x+6}}dx$

Complete square:

$4\int\frac{1}{\sqrt{(\frac{2x+3}{2})^{2}+\frac{15} {4}}dx$

Let $u=\frac{2x+3}{2}, \;\ du=dx$

Then, we get:

$2\int\frac{1}{\sqrt{4u^{2}-15}}du$

Now, try a trig sub perhaps to finish:

$2\int\frac{1}{\sqrt{4u^{2}+15}}du$

Now, let $u=\frac{\sqrt{15}}{2}tan(t), \;\ du=\frac{\sqrt{15}}{2}sec^{2}(t)dt$

Make the subs and it becomes:

$\int sec(t)dt$

Integrate and make the back substitutions. The solution involves ln

gashan · Sep 25, 2010, 06:28 AM

Quote:

Originally Posted by galactus

No, it is a bit more involved than that.

We start with:

$4\int\frac{1}{\sqrt{x^{2}+3x+6}}dx$

Complete square:

$4\int\frac{1}{\sqrt{(\frac{2x+3}{2})^{2}+\frac{15} {4}}dx$

Let $u=\frac{2x+3}{2}, \;\ du=dx$

Then, we get:

$2\int\frac{1}{\sqrt{4u^{2}-15}}du$

Now, try a trig sub perhaps to finish:

$2\int\frac{1}{\sqrt{4u^{2}+15}}du$

Now, let $u=\frac{\sqrt{15}}{2}tan(t), \;\ du=\frac{\sqrt{15}}{2}sec^{2}(t)dt$

Make the subs and it becomes:

$\int sec(t)dt$

Integrate and make the back substitutions. The solution involves ln

ok cool. Thank you. Help greatly appreciated.

gashan · Sep 25, 2010, 11:27 AM

Can you confirm if these answers that we got are correct. We placed x=0 in the second question, and both sides equate to the same. But just wanted confirmation.
' (2x + 5)/(x + 3x + 6) dx
= ' (2x + 3 + 2)/(x + 3x + 6) dx
= ' (2x + 3)/(x + 3x + 6) dx + 2 ' dx/(x + 3x + 6)
= ' (2x + 3)/(x + 3x + 6) dx + 2 ' dx/((x + 3/2) + 15/4)
= ' (2x + 3)/(x + 3x + 6) dx + 2 ' dx/((x + 3/2) + '(15)/2)
= ln (x + 3x + 6) + 2 (2/'15) arctan( 2(x + 3/2)/'15) + C
= ln (x + 3x + 6) + (4/'15) arctan( 2(x + 3/2)/'15) + C

' (x + 5x + 4)/[x(x + 3x + 2) ] dx
= ' (x + 5x + 4)/(x + 3x + 2x) dx
= ' (x + 3x + 2x + 2x - 2x + 4)/(x + 3x + 2x) dx
= ' (x + 3x + 2x)/(x + 3x + 2x) dx + ' (2x - 2x + 4)/(x + 3x + 2x) dx

Let
(2x - 2x + 4)/(x + 3x + 2x) = A/x + B/(x + 1) + C/(x + 2)
Then
2x - 2x + 4 = A(x + 1)(x + 2) + Bx(x + 2) + Cx(x + 1)
2x - 2x + 4 = x(A + B + C) + x(3A + 2B + C) + (2A)

By comparing coefficients of different powers of x:
2A = 4 ==> A = 2

-2 = 3A + 2B + C ==> -2 = 6 + 2B + C ==> 2B + C = -8

2 = A + B + C ==> 2 = 2 + B + C ==> B + C = 0 ==> B = -C

==> -2C + C = -8 ==> C = 8 ==> B = 8

==> (2x - 2x + 4)/(x + 3x + 2x) = 2/x + 8/(x + 1) - 8/(x + 2)

Continuation
= ' (x + 3x + 2x)/(x + 3x + 2x) dx + ' (2x - 2x + 4)/(x + 3x + 2x) dx
= ' 1 dx + ' (2/x + 8/(x + 1) - 8/(x + 2)) dx
= x + 2ln|x| + 8ln|x + 1| - 8ln|x + 2| + C

Unknown008 · Sep 25, 2010, 12:03 PM

1. $\int \frac{2x+5}{x^2 + 3x + 6} dx = ln|x^2 + 3x + 6| + 2\int \frac{1}{x^2 + 3x + 6} dx$

Complete the square;

$= \ln|x^2 + 3x + 6| + 2\int \frac{1}{$x+\frac32$^2 + \frac{15}{4}} dx$

Sub you = x + 3/2.

Du = dx

$= \ln|x^2 + 3x + 6| + 2\int \frac{1}{u^2 + \frac{15}{4}} du$

Use $u = \frac{\sqrt{15}}{2}tan(t)$

$du = \frac{15}{4}sec^2(t) dt$

$= \ln|x^2 + 3x + 6| + 2\int \frac{1}{\frac{15}{4}tan^2(t) + \frac{15}{4}} . \frac{\sqrt{15}}{2}sec^2(t) dt$

$= \ln|x^2 + 3x + 6| + 2\int \frac{2}{\sqrt{15}} dt$

$= \ln|x^2 + 3x + 6| + 2$\frac{2t}{\sqrt{15}}$ +c$

$= \ln|x^2 + 3x + 6| + \frac{4t}{\sqrt{15}}+c$

-> $t = tan^{-1}$\frac{2u}{\sqrt{15}}$+c$

$= \ln|x^2 + 3x + 6| + \frac{4}{\sqrt{15}}tan^{-1}$\frac{2u}{\sqrt{15}}$+c$

$= \ln|x^2 + 3x + 6| + \frac{4tan^{-1}$\frac{2x + 3}{\sqrt{15}}$}{\sqrt{15}}+c$

Okay, seems it agrees with your answer, good

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