Understanding Engineering Mathematics

Solution to review question 9.1.7

If, in these questions, you have done something like:

‘

∫

dx

x^2 + 2 x+ 2

=

1

2 x+ 2

ln(x^2 + 2 x+ 2 )’

which isincorrect, then you have probably misunderstood the idea of

substitution. ‘Dividing by the derivative of the denominator’ in this

way willonlywork for a linear denominator, when the derivative is

constant (261

➤

). You have only to differentiate the right-hand side

to see that it cannot possibly be the integral of the left-hand side. The

following solutions show the correct approach to such integrals.

A.In this case the denominator factorises and we don’t need to complete

the square – we can split into partial fractions:

∫

dx

x^2 +x− 2

=

∫

dx

(x− 1 )(x+ 2 )

=

∫ [

1

3 (x− 1 )

−

1

3 (x+ 2 )

]

dx

=

1

3

∫

dx

x− 1

−

1

3

∫

dx

x+ 2

+C

=

1

3

ln(x− 1 )−

1

3

ln(x+ 2 )+C

using ∫

dx

x−a

=ln(x−a)+C

We can tidy the answer up and write:

∫

dx

x^2 +x− 2

=

1

3

ln

(

x− 1

x+ 2

)

+C

B.In this case the denominator does not factorise, and the only option is

to complete the square (66

➤

). We have:

∫

dx

x^2 + 2 x+ 2

=

∫

dx

(x+ 1 )^2 + 1

Now this looks like the inverse tan integral (256

➤

), which we can

get by substitutingu=x+1,du=dx:

∫

dx

(x+ 1 )^2 + 1

=

∫

du

u^2 + 1

=tan−^1 u

so ∫

dx

x^2 + 2 x+ 2

=tan−^1 (x+ 1 )