Understanding Engineering Mathematics

Solution to review question 9.1.6

(i) In the integral

∫

x+ 1

x^2 + 2 x+ 3

dxthe derivative ofx^2 + 2 x+3is

2 (x+ 1 )and we have anx+1 on the top. This suggests substituting

u=x^2 + 2 x+3. Then

du

dx

= 2 (x+ 1 )

You may now be tempted to write

dx=

du

2 (x+ 1 )

substitute in the integral and cancel thex+1. While correct, this is

a bad habit – never mix variables in an integral. Instead, simply use

the fact thatx+1 is ready and waiting for us in the integrand and

write

du= 2 (x+ 1 )dx

or

(x+ 1 )dx=

du

2

to get ∫

(x+ 1 )dx

x^2 + 2 x+ 3

=

∫

1

u

du

2

=

1

2

∫

du

u

=

1

2

lnu+C

=

1

2

ln(x^2 + 2 x+ 3 )+C

(ii) Again, in

∫

xsin(x^2 + 1 )dxthe derivative ofx^2 +1is2xand we

have anx multiplying the sine, which suggests puttingu=x^2 +

1. Then ∫
   xsin(x^2 + 1 )dx−−−→

∫

sinu

1

2

du

=

1

2

∫

sinudu

=−

1

2

cosu+C

=−

1

2

cos(x^2 + 1 )+C