CBSE Board Examination 2023 · Section C · 3 marks

Q27: $\int\frac{dx}{\sqrt x(\sqrt x + 1)(\sqrt x + 2)}$

The question

Find :

\int\frac{1}{\sqrt x(\sqrt x + 1)(\sqrt x + 2)}\,dx

Exam answer

Let $I = \displaystyle\int\frac{dx}{\sqrt x(\sqrt x + 1)(\sqrt x + 2)}$ ⁠.

Let $\sqrt x = t$ ⁠, $\frac{1}{2\sqrt x}\,dx = dt$ ⁠.

Why?

$\sqrt x$ appears in every factor of the denominator, so $t = \sqrt x$ turns everything into $t$ ⁠. Its differential comes from

\frac{d}{dx}\sqrt x = \frac{1}{2\sqrt x}

and the integrand already has $\frac{1}{\sqrt x}$ ⁠, which is twice this.

Multiply both sides of $\frac{1}{2\sqrt x}\,dx = dt$ by 2:

\begin{aligned} 2 \times \frac{dx}{2\sqrt x} &= 2\,dt \\ \frac{dx}{\sqrt x} &= 2\,dt \end{aligned}

\Rightarrow\ \frac{dx}{\sqrt x} = 2\,dt

\therefore\ I = 2\int\frac{dt}{(t + 1)(t + 2)}

Why?

The factor $\frac{1}{\sqrt x}$ goes with $dx$ to make $2\,dt$ ⁠. The brackets become $t + 1$ and $t + 2$ ⁠.

\begin{aligned} I &= \int\frac{1}{(\sqrt x + 1)(\sqrt x + 2)} \cdot \frac{dx}{\sqrt x} \\ &= \int\frac{1}{(t + 1)(t + 2)} \cdot 2\,dt \\ &= 2\int\frac{dt}{(t + 1)(t + 2)} \end{aligned}

Let $\frac{1}{(t + 1)(t + 2)} = \frac{A}{t + 1} + \frac{B}{t + 2}$ ⁠.

Why?

The denominator has two different linear factors, so it splits into partial fractions with one constant over each factor.

Then $1 = A(t + 2) + B(t + 1)$ ⁠. Putting $t = -1$ ⁠: $A = 1$ ⁠. Putting $t = -2$ ⁠: $B = -1$ ⁠.

Why?

Multiply both sides by $(t + 1)(t + 2)$ to clear the fractions. Then choose values of $t$ that make one bracket zero:

Each term times $(t + 1)(t + 2)$ ⁠: the left side leaves 1, $\frac{A}{t + 1}$ leaves $A(t + 2)$ ⁠, and $\frac{B}{t + 2}$ leaves $B(t + 1)$ ⁠.

\begin{aligned} &t = -1: \\ &1 = A(-1 + 2) + B(-1 + 1) \\ &1 = A(1) + B(0) \\ &A = 1 \\[4pt] &t = -2: \\ &1 = A(-2 + 2) + B(-2 + 1) \\ &1 = A(0) + B(-1) \\ &1 = -B \\ &B = -1 \end{aligned}

So $\frac{A}{t + 1} + \frac{B}{t + 2} = \frac{1}{t + 1} - \frac{1}{t + 2}$ ⁠.

\begin{aligned} I &= 2\int\left(\frac{1}{t + 1} - \frac{1}{t + 2}\right)dt \\ &= 2\big[\log|t + 1| - \log|t + 2|\big] + C \end{aligned}

Why?

Standard integral:

\int\frac{dt}{t + k} = \log|t + k| + C

Use it with $k = 1$ and with $k = 2$ ⁠. One constant $C$ is enough for the whole answer.

\begin{aligned} I &= 2\big[\log(\sqrt x + 1) \\ &\qquad - \log(\sqrt x + 2)\big] + C \end{aligned}

Why?

Put back $t = \sqrt x$ ⁠. Since $\sqrt x \ge 0$ ⁠, both $\sqrt x + 1$ and $\sqrt x + 2$ are positive, so the modulus signs can go.

Then $\log m - \log n = \log\frac{m}{n}$ gives the single log in the answer.

\begin{aligned} &\log(\sqrt x + 1) - \log(\sqrt x + 2) \\ &= \log\frac{\sqrt x + 1}{\sqrt x + 2} \end{aligned}

The 2 in front stays in front.

$\therefore\ I = 2\log\left(\dfrac{\sqrt x + 1}{\sqrt x + 2}\right) + C$ ⁠.

Where this comes from

Chapter 7, Integrals: Substitution, then partial fractions.

Question type: Substitution + Partial fractions (combines 2 types)

How to spot it: click to see

Type 2. Substitution: when the derivative is hiding in the question

Spot it: a function and its derivative both appear, like $\log x$ and $\frac{1}{x}$ ⁠, or $x^2 + 1$ and $x$ ⁠.

Here: First, $\sqrt x$ is in every bracket and $\frac{1}{\sqrt x}$ sits beside $dx$ ⁠: put $\sqrt x = t$ ⁠, so $\frac{dx}{\sqrt x} = 2\,dt$ ⁠.

Type 7. Partial fractions: break a fraction into simple pieces

Spot it: a fraction of two polynomials whose bottom splits into factors.

Here: The fraction left in $t$ has the bottom $(t + 1)(t + 2)$ ⁠: split it into $\frac{1}{t + 1} - \frac{1}{t + 2}$ ⁠, which gives two logs.

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Integrals in past papers: question 16 of 46 · see them all

Q27: ∫dxx(x+1)(x+2)\int\frac{dx}{\sqrt x(\sqrt x + 1)(\sqrt x + 2)}∫x​(x​+1)(x​+2)dx​