CBSE Sample Paper 2024-25 · Section C · 3 marks

Q29: $\int\left\{\frac{1}{\log x} - \frac{1}{(\log x)^2}\right\}dx$ ⁠, or $\int_0^1 x(1 - x)^n\,dx$

Sample 2024-25Ch 7 · Integrals

The question

(a) Evaluate: $\displaystyle\int\left\{\frac{1}{\log x} - \frac{1}{(\log x)^2}\right\}dx$ ⁠; (where $x > 1$ ⁠).

(b) Evaluate : $\displaystyle\int_0^1 x(1 - x)^n\,dx$ ⁠; (where $n \in N$ ⁠).

Exam answer

(a)

Let $I = \displaystyle\int\left\{\frac{1}{\log x} - \frac{1}{(\log x)^2}\right\}dx$

I = \int\frac{dx}{\log x} - \int\frac{dx}{(\log x)^2}

Why?

Neither $\frac{1}{\log x}$ nor $\frac{1}{(\log x)^2}$ has an antiderivative we can write down on its own. The plan is to integrate the first one by parts and watch the second one cancel.

Integrating the first integral by parts, with $\frac{1}{\log x}$ as the first function and 1 as the second,

Why?

Integration by parts:

\begin{aligned} \int u\,v\,dx &= u\int v\,dx \\ &\quad - \int\left(\frac{du}{dx}\int v\,dx\right)dx \end{aligned}

Writing $\frac{1}{\log x} = \frac{1}{\log x}\cdot 1$ lets us use it with $v = 1$ ⁠, which is easy to integrate, while $\frac{1}{\log x}$ is easy to differentiate.

\begin{aligned} I &= \frac{1}{\log x}\int dx \\ &\quad - \int\left\{\frac{d}{dx}\left(\frac{1}{\log x}\right)\int dx\right\}dx \\ &\quad - \int\frac{dx}{(\log x)^2} \end{aligned}

Why?

$\int dx = x$ ⁠. Write $\frac{1}{\log x} = (\log x)^{-1}$ and use the chain rule:

\begin{aligned} &\frac{d}{dx}(\log x)^{-1} \\ &= -(\log x)^{-2}\cdot\frac1x \\ &= -\frac{1}{x(\log x)^2} \end{aligned}

\begin{aligned} &= \frac{x}{\log x} + \int\frac{1}{(\log x)^2}\cdot\frac1x\cdot x\,dx \\ &\quad - \int\frac{dx}{(\log x)^2} \end{aligned}

Why?

The minus in the formula and the minus in the derivative give a plus.

\begin{aligned} &-\int\left\{-\frac{1}{x(\log x)^2}\cdot x\right\}dx \\ &= +\int\frac{1}{x(\log x)^2}\cdot x\,dx \\ &= \int\frac{1}{(\log x)^2}\cdot\frac1x\cdot x\,dx \end{aligned}

The first part is $\frac{1}{\log x}\cdot x = \frac{x}{\log x}$ ⁠.

Inside the integral, $\frac1x\cdot x = 1$ ⁠, so the integral is $\int\frac{dx}{(\log x)^2}$ ⁠.

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

Why?

The two remaining integrals are the same, so they cancel. Only one arbitrary constant $c$ is needed at the end.

Check: differentiating $\frac{x}{\log x}$ by the quotient rule gives back the integrand:

With $u = x$ ⁠, $v = \log x$ ⁠, $u' = 1$ ⁠, $v' = \frac1x$ ⁠:

\begin{aligned} \frac{d}{dx}\left(\frac{x}{\log x}\right) &= \frac{u'v - uv'}{v^2} \\ &= \frac{(1)\log x - x\cdot\frac1x}{(\log x)^2} \\ &= \frac{\log x - 1}{(\log x)^2} \\ &= \frac{\log x}{(\log x)^2} - \frac{1}{(\log x)^2} \end{aligned}

\frac{\log x - 1}{(\log x)^2} = \frac{1}{\log x} - \frac{1}{(\log x)^2}

$\therefore\ I = \dfrac{x}{\log x} + c$ , where $c$ is an arbitrary constant of integration.

(b)

Let $I = \displaystyle\int_0^1 x(1 - x)^n\,dx$ ⁠.

Using $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ ⁠,

\begin{aligned} I &= \int_0^1 (1 - x)\{1 - (1 - x)\}^n\,dx \\ &= \int_0^1 x^n(1 - x)\,dx \end{aligned}

Why?

Expanding $(1 - x)^n$ for a general $n$ would be very long. The property moves the power onto a single $x$ instead.

Here $a = 1$ ⁠, so every $x$ is replaced by $1 - x$ ⁠:

1 - (1 - x) = x

So $(1 - x)\{1 - (1 - x)\}^n$ becomes $(1 - x)x^n$ ⁠.

\begin{aligned} I &= \int_0^1 x^n\,dx - \int_0^1 x^{n + 1}\,dx \\ &= \frac{1}{n + 1}\Big[x^{n + 1}\Big]_0^1 - \frac{1}{n + 2}\Big[x^{n + 2}\Big]_0^1 \end{aligned}

Why?

Multiply out: $x^n(1 - x) = x^n - x^{n + 1}$ ⁠.

Then use the power rule for integrals:

\int x^m\,dx = \frac{x^{m + 1}}{m + 1}

with $m = n$ and with $m = n + 1$ ⁠.

\begin{aligned} I &= \frac{1}{n + 1} - \frac{1}{n + 2} \\ &= \frac{(n + 2) - (n + 1)}{(n + 1)(n + 2)} \end{aligned}

Why?

At the upper limit $1^{n + 1} = 1^{n + 2} = 1$ ⁠. At the lower limit both powers of 0 are 0, since $n \ge 1$ ⁠.

\begin{aligned} \Big[x^{n + 1}\Big]_0^1 &= 1^{n + 1} - 0^{n + 1} \\ &= 1 - 0 = 1 \\ \Big[x^{n + 2}\Big]_0^1 &= 1^{n + 2} - 0^{n + 2} \\ &= 1 - 0 = 1 \end{aligned}

So $I = \frac{1}{n + 1}(1) - \frac{1}{n + 2}(1)$ ⁠.

Take the common denominator $(n + 1)(n + 2)$ ⁠. The numerator is $n + 2 - n - 1 = 1$ ⁠.

\begin{aligned} \frac{1}{n + 1} &= \frac{n + 2}{(n + 1)(n + 2)} \\ \frac{1}{n + 2} &= \frac{n + 1}{(n + 1)(n + 2)} \\[4pt] (n + 2) - (n + 1) &= n + 2 - n - 1 \\ &= (n - n) + (2 - 1) \\ &= 0 + 1 = 1 \end{aligned}

$\therefore\ I = \dfrac{1}{(n + 1)(n + 2)}$ ⁠.

Where this comes from

Chapter 7, Integrals: Integration by parts; definite integral property.

Question type: (a) Integration by parts; (b) Limits from 0 to $a$

How to spot it: click to see

(a) Type 8. Integration by parts: a product of two different kinds

Spot it: a product such as $x\sin x$ ⁠, $xe^x$ or $x\log x$ ⁠. A lone $\log x$ or $\tan^{-1}x$ works too: take the other function as 1.

Here: Neither $\frac{1}{\log x}$ nor $\frac{1}{(\log x)^2}$ has a card. Integrate $\frac{1}{\log x}$ by parts with 1 as the second function: it leaves $+\int\frac{dx}{(\log x)^2}$ ⁠, which cancels the second term.

(b) Type 12. Limits from 0 to $a$ ⁠: the $f(a - x)$ trick

Spot it: replacing $x$ by $a - x$ swaps the parts of the question, like sin and cos when $a$ is $\frac{\pi}{2}$ ⁠.

Here: With $a = 1$ ⁠, replacing $x$ by $1 - x$ swaps $x$ and $1 - x$ ⁠, so the power $n$ moves onto a plain $x$ ⁠. The integral becomes $\int_0^1(x^n - x^{n + 1})\,dx$ ⁠, which the power card finishes.

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Integrals in past papers: question 31 of 46 · see them all

Q29: ∫{1log⁡x−1(log⁡x)2}dx\int\left\{\frac{1}{\log x} - \frac{1}{(\log x)^2}\right\}dx∫{logx1​−(logx)21​}dx⁠, or ∫01x(1−x)n dx\int_0^1 x(1 - x)^n\,dx∫01​x(1−x)ndx

Q29: $\int\left\{\frac{1}{\log x} - \frac{1}{(\log x)^2}\right\}dx$ ⁠, or $\int_0^1 x(1 - x)^n\,dx$