Understanding Engineering Mathematics

C.This question illustrates that sometimes you may need to combine
methods. Here, the numerator is of the same degree as the denominator,
so we must first divide out:
2 x^2 + 5 x+ 4
x^2 + 2 x+ 2

=

2 (x^2 + 2 x+ 2 )− 4 x− 4 + 5 x+ 4

x^2 + 2 x+ 2

= 2 +

x

x^2 + 2 x+ 2

So ∫

2 x^2 + 5 x+ 4

x^2 + 2 x+ 2

dx=

∫ [

2 +

x

x^2 + 2 x+ 2

]

dx

= 2 x+C+

∫

xdx

x^2 + 2 x+ 2

In the remaining integral the denominator doesn’t factorise and the top

is not the derivative of the bottom. We can remedy this by adding and

subtracting 1 in the numerator, to manufacture the derivative of the

bottom on the top!

∫

xdx

x^2 + 2 x+ 2

=

∫

x+ 1 − 1

x^2 + 2 x+ 2

dx

=

∫

(x+ 1 )

x^2 + 2 x+ 2

dx−

∫

dx

x^2 + 2 x+ 2

=

1

2

∫

d(x^2 + 2 x+ 2 )

x^2 + 2 x+ 2

−

∫

d(x+ 1 )

(x+ 1 )^2 + 1

=

1

2

ln|x^2 + 2 x+ 2 |−tan−^1 (x+ 1 )+C

So, finally, the integral is

2 x+

1

2

ln|x^2 + 2 x+ 2 |−tan−^1 (x+ 1 )+C

Note that we don’t need to duplicate the arbitrary constantC.
D.In these problems you have to decide which method to use for yourself.
Also, (iv) illustrates that the ‘obvious’ method is not always the best,
so you should always be on the look-out for easier alternatives.
(i) The denominator won’t factorise, so complete the square
∫
dx
x^2 +x+ 1

=

∫

dx

(

x+

1

2

) 2

+

3

4

=

2

√

3

tan−^1







x+

1

√^2

3 / 2





+C

=

2

√

3

tan−^1

(

2 x+ 1

√

3

)

+C