2022 H2 Maths Paper 1 Q8 — Worked Solution

Part (a)

4 marks

Approach

the denominator factorises as a perfect square, so the substitution $u = x + 1$ turns the integral into a sum of two standard forms.

\int\frac{2x - 1}{x^{2} + 2x + 1}\, dx = \int\frac{2x - 1}{(x + 1)^{2}}\, dx

Using the substitution $u = x + 1$ , so $x = u - 1$ and $dx = du$ , the numerator becomes

2x - 1

= 2(u - 1) - 1

= 2u - 3

\int\frac{2u - 3}{u^{2}}\, du

= \int\left( \frac{2}{u} - \frac{3}{u^{2}} \right)du

= 2\ln|u| + \frac{3}{u} + c

Substituting back $u = x + 1$ :

\int\frac{2x - 1}{x^{2} + 2x + 1}\, dx = 2\ln|x + 1| + \frac{3}{x + 1} + c

To check your answer: differentiating,

\frac{d}{dx}\left\lbrack 2\ln(x + 1) + \frac{3}{x + 1} \right\rbrack = \frac{2}{x + 1} - \frac{3}{(x + 1)^{2}} = \frac{2x - 1}{(x + 1)^{2}}

✓

$\therefore$ $\int\frac{2x - 1}{x^{2} + 2x + 1}\, dx = 2\ln|x + 1| + \frac{3}{x + 1} + c$ .

3 marks

Approach

the modulus forces us to split the interval where $2x - 1$ changes sign, at $x = \frac{1}{2}$ , and reuse the antiderivative from (a) on each piece.

Since $2x - 1 = 0$ at $x = \frac{1}{2}$ : for $0 \leq x < \frac{1}{2}$ , $2x - 1 < 0$ ; for $\frac{1}{2} < x \leq 2$ , $2x - 1 > 0$ . So

\int_{0}^{2}\frac{|2x - 1|}{(x + 1)^{2}}\, dx = \int_{0}^{1/2} - \frac{2x - 1}{(x + 1)^{2}}\, dx + \int_{1/2}^{2}\frac{2x - 1}{(x + 1)^{2}}\, dx

Let $F(x) = 2\ln(x + 1) + \frac{3}{x + 1}$ ; the modulus signs from part (a) can be dropped since $x + 1 > 0$ throughout $\lbrack 0,2\rbrack$ . Then

\int_{0}^{2}\frac{|2x - 1|}{(x + 1)^{2}}\, dx = \left\lbrack F(0) - F\left( \frac{1}{2} \right) \right\rbrack + \left\lbrack F(2) - F\left( \frac{1}{2} \right) \right\rbrack

= F(0) + F(2) - 2F\left( \frac{1}{2} \right)

Evaluating $F$ at each point:

F(0)

= 2\ln 1 + 3

= 3

F\left( \frac{1}{2} \right)

= 2\ln\frac{3}{2} + 2

F(2)

= 2\ln 3 + 1

\int_{0}^{2}\frac{|2x - 1|}{(x + 1)^{2}}\, dx

= 3 + \left( 2\ln 3 + 1 \right) - 2\left( 2\ln\frac{3}{2} + 2 \right)

= 2\ln 3 - 4\ln\frac{3}{2}

Using $\ln\frac{3}{2} = \ln 3 - \ln 2$ :

2\ln 3 - 4\ln\frac{3}{2}

= 2\ln 3 - 4\left( \ln 3 - \ln 2 \right)

= 4\ln 2 - 2\ln 3

= 2\ln\frac{4}{3}

To check your answer:

2\ln\frac{4}{3} \approx 0.575

, matching a numerical estimate of

\int_{0}^{2}\frac{|2x - 1|}{(x + 1)^{2}}\, dx

✓

$\therefore$ $\int_{0}^{2}\frac{|2x - 1|}{x^{2} + 2x + 1}\, dx = 2\ln\frac{4}{3}$ .