CBSE Sample Paper 2021-22 (Term 2) · Section B · 3 marks

Q7: $\int \frac{x + 1}{(x^2 + 1)x}\,dx$

Sample 2021-22 T2Ch 7 · Integrals

The question

Find: $\displaystyle\int \frac{x + 1}{(x^2 + 1)x}\,dx$

Exam answer

Let $\frac{x + 1}{(x^2 + 1)x} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x}$ ⁠.

Why?

The denominator has a linear factor $x$ and a quadratic factor $x^2 + 1$ ⁠, which cannot be factorised because it has no real roots.

A linear factor gets a constant on top. A quadratic factor that cannot be factorised gets $Ax + B$ on top.

\begin{aligned} \Rightarrow x + 1 &= (Ax + B)x \\ &\quad + C(x^2 + 1) \end{aligned}

Why?

Multiply both sides by $(x^2 + 1)x$ ⁠. The result is an identity: it is true for every value of $x$ ⁠.

Equating the coefficients of $x^2$ ⁠, $x$ and the constant terms,

A + C = 0,\quad B = 1,\quad C = 1

Why?

Expand the right side:

\begin{aligned} &(Ax + B)x + C(x^2 + 1) \\ &= Ax^2 + Bx + Cx^2 + C \\ &= (A + C)x^2 + Bx + C \end{aligned}

The left side is $0x^2 + 1x + 1$ ⁠. Matching the coefficients gives the three equations. Then $A = -C = -1$ ⁠.

\begin{aligned} x^2&:\ A + C = 0 \\ x&:\ B = 1 \\ \text{constant}&:\ C = 1 \\[4pt] A + 1 &= 0 \Rightarrow A = -1 \end{aligned}

\therefore\ A = -1,\ B = 1,\ C = 1

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

\begin{aligned} I &= \int \frac{-x + 1}{x^2 + 1}\,dx + \int \frac{dx}{x} \\ &= -\frac12\int \frac{2x - 2}{x^2 + 1}\,dx + \int \frac{dx}{x} \end{aligned}

Why?

With $A = -1$ ⁠, $B = 1$ ⁠, $C = 1$ ⁠:

\begin{aligned} \frac{Ax + B}{x^2 + 1} &= \frac{-x + 1}{x^2 + 1} \\ \frac{C}{x} &= \frac1x \end{aligned}

The derivative of $x^2 + 1$ is $2x$ ⁠. To make $2x$ appear on top, write $-x + 1 = -\frac12(2x - 2)$ ⁠.

\begin{aligned} -\frac12(2x - 2) &= -\frac12(2x) - \frac12(-2) \\ &= -x + 1 \end{aligned}

Then split the fraction:

\begin{aligned} &\frac{2x - 2}{x^2 + 1} \\ &= \frac{2x}{x^2 + 1} - \frac{2}{x^2 + 1} \end{aligned}

and $-\frac12 \times (-2) = +1$ ⁠.

\begin{aligned} I &= -\frac12\int \frac{2x}{x^2 + 1}\,dx \\ &\quad + \int \frac{dx}{x^2 + 1} + \int \frac{dx}{x} \end{aligned}

Why?

Standard integrals:

\begin{aligned} \int \frac{f'(x)}{f(x)}\,dx &= \log|f(x)| + C \\ \int \frac{dx}{1 + x^2} &= \tan^{-1}x + C \end{aligned}

Since $x^2 + 1 > 0$ for every $x$ ⁠, $\log(x^2 + 1)$ needs no modulus. $\log|x|$ does.

Use them on each piece. In the first, $f(x) = x^2 + 1$ and $f'(x) = 2x$ ⁠:

\begin{aligned} -\frac12\int \frac{2x}{x^2 + 1}\,dx &= -\frac12\log(x^2 + 1) \\ \int \frac{dx}{x^2 + 1} &= \tan^{-1}x \\ \int \frac{dx}{x} &= \log|x| \end{aligned}

Add the three and one constant $c$ ⁠.

$\therefore\ I = -\dfrac12\log(x^2 + 1) + \tan^{-1}x$ $+ \log|x| + c$

Where this comes from

Chapter 7, Integrals: Partial fractions with a quadratic factor.

Question type: Partial fractions + A linear top over a quadratic (combines 2 types)

How to spot it: click to see

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

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

Here: First, the bottom $(x^2 + 1)x$ splits into $x$ and $x^2 + 1$ ⁠, which cannot be factorised. So write $\frac{Ax + B}{x^2 + 1} + \frac{C}{x}$ and compare coefficients: $A = -1$ ⁠, $B = 1$ ⁠, $C = 1$ ⁠.

Type 5. A linear top over a quadratic: split the top

Spot it: $px + q$ on top, and a quadratic, or the square root of one, at the bottom.

Here: The piece left, $\frac{-x + 1}{x^2 + 1}$ ⁠, has a linear top over a quadratic. Write $-x + 1$ as $-\frac12(2x - 2)$ ⁠: the $2x$ part gives $\log(x^2 + 1)$ ⁠, and the number part gives $\tan^{-1}x$ ⁠.

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