# Need help evaluating this integral: `int(2x+3)/(4x^2+4x+5)dx` I completed the square and got:`int(2x+3)/(4(x+1/2)^2+4)dx` ` ` This problem looks like a tangent problem so I did: u=atan`theta` This...

Need help evaluating this integral: `int(2x+3)/(4x^2+4x+5)dx`

I completed the square and got:`int(2x+3)/(4(x+1/2)^2+4)dx` ` `

This problem looks like a tangent problem so I did: u=atan`theta`

This gave then turned into: `2(x+1/2)^2=2tantheta`

Now, what's bothering me is the 2x+3 in the numerator, how do I cancel that out?

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`int(2x+3)/(4x^2+4x+5) dx`

Since there is x variable at the numerator, to solve it, apply u-substitution method. So,

`u = 4x^2+4x +5`

Then, differentiate it with respect to x.

`du = (8x + 4)dx`

d`u=4(2x+ 1)dx`

`(du)/4=(2x+ 1)dx`

Notice that in our integrand, we have 2x + 3. So, to be able to apply u-substitution method, express the integrand as sum of two fractions.

`int (2x+3)/(4x^2+4x+5)dx`

`= int [(2x + 1)/(4x^2+4x+5)+2/(4x^2+4x+5)]dx`

`=int (2x+ 1)/(4x^2+4x+5)dx + int 2/(4x^2+4x+5)dx`

So for our first integral, let's proceed to apply the u-substitution u = 4x^2 +4x+5 and du/4=(2x + 1)dx.

`=int 1/u*(du)/4 + int2/(4x^2+4x+5)dx`

`=int (du)/(4u) + int2/(4x^2+4x+5)dx`

Then, apply the formula int 1/u du = ln |u| + C.

`= (ln |u|)/4+C+ int 2/(4x^2 + 4x + 5)dx`

And, substitute back u = 4x^2+4x + 5.

`= (ln|4x^2+4x+5|)/4+C + int2/(4x^2+4x+5)dx`

For the second integral, it no longer has x at the numerator. To evaluate it, we have to do the completing the square method at the denominator. So, second integral becomes:

`=(ln|4x^2+4x+5|)/4+C + int 2/(4(x+1/2)^2 + 4)dx`

`=(ln|4x^2+4x+5|)/4+C +int2/(4[(x+1/2)^2+1])dx`

`=(ln|4x^2+4x+5|)/4+C +1/2int 1/((x+1/2)^2+1)dx`

Then, apply the formula int 1/(a^2+u^2)du= 1/atan^(-1)(u/a) +C. So in our second integral, we have:

u = x + 1/2

du = dx

a = 1

Substituting them to the formula yields,

`=(ln|4x^2+4x+5|)/4+C + 1/2*1/1tan^(-1)( (x+1/2)/1)+C`

Since C represents any number, we can express it with single C only.

`=(ln|4x^2+4x+5|)/4+1/2tan^(-1)(x+1/2) + C`

**Therefore, `int(2x+3)/(4x^2+4x+5)dx=(ln|4x^2+4x+5|)/4+1/2tan^(-1)(x+1/2) + C` .**