Understanding Engineering Mathematics

(ii) The denominator factorises so we can use partial fractions

∫

dx

x^2 + 3 x+ 2

=

∫

dx

(x+ 1 )(x+ 2 )

=

∫ [

1

(x+ 1 )

−

1

(x+ 2 )

]

dx

=

∫

dx

x+ 1

−

∫

dx

x+ 2

=ln

(

x+ 1

x+ 2

)

+C

(iii) The top is the derivative of bottom so use substitution

∫

2 x+ 1

x^2 +x− 1

dx=

∫

d(x^2 +x− 1 )

x^2 +x− 1

=ln(x^2 +x− 1 )+C

(iv) You may have tackled this by partial fractions. This is valid and

will give the correct answer, but you might spot an easier way, if

your algebra and substitution are on the ball.

∫

3 x

(x− 1 )(x+ 1 )

dx= 3

∫

x

x^2 − 1

dx

=

3

2

∫

2 x

x^2 − 1

dx

=

3

2

∫

d(x^2 )

x^2 − 1

=

3

2

ln(x^2 − 1 )+C

9.2.8 Using trig identities in integration

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252 283➤

Using a simple linear substitution,u= 3 x, we can integrate things such as:

∫

cos 3xdx=

1

3

sin 3x+C

Using this and various trig identities we can perform some quite complicated integrations.
The most common examples here include the use of the compound angle and double angle
formulae

sin(A±B)=sinAcosB±sinBcosA

cos(A±B)=cosAcosB∓sinAsinB