(a) Find: ∫ (cos x)/((2 + sin x)(4 + sin x)) dx OR (b) Find: ∫ (x + 3)/(x² + 4x + 5) dx

Question

\int \frac{\cos x}{(2 + \sin x)(4 + \sin x)}\, dx

(OR)

(b) Find:

\int \frac{x + 3}{x^2 + 4x + 5}\, dx

CBSECBSE Class XII Board 2026Subjective· 3mImportance★★★★★

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Part (a)Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

Adding $\frac{2}{x-1}+\frac{3}{x+2}$ over a common denominator gives $\frac{5x+1}{x^2+x-2}$ . Partial fraction decomposition reverses this — it breaks one complicated rational function back into a sum of simple pieces. Those pieces are far easier to integrate: $\int \frac{2}{x-1}\,dx = 2\log|x-1|$ is immediate, while the combined fraction is not.

When it applies

You need a proper rational function, $\deg P < \deg Q$ . If the numerator's degree is equal or higher, first do polynomial long division. The denominator $Q(x)$ must be factored into linear and/or irreducible quadratic factors.

The standard forms

For $\frac{P(x)}{Q(x)}$ with $Q$ fully factored, each factor contributes a term:

Distinct linear $(ax+b)$ $\to$ $\dfrac{A}{ax+b}$ .
Repeated linear $(ax+b)^n$ $\to$ $\dfrac{A_1}{ax+b}+\dfrac{A_2}{(ax+b)^2}+\cdots+\dfrac{A_n}{(ax+b)^n}$ .
Irreducible quadratic $(ax^2+bx+c)$ $\to$ $\dfrac{Ax+B}{ax^2+bx+c}$ — a linear numerator, not just a constant.

The unknown constants are found from the resulting equations; the decomposition is unique, which is what lets us solve for them systematically.

Worked example

Decompose $\frac{3x+5}{x^2-3x+2}$ . Factor the denominator: $(x-1)(x-2)$ . Set

\frac{3x+5}{(x-1)(x-2)} = \frac{A}{x-1}+\frac{B}{x-2}.

Clear denominators: $3x+5 = A(x-2)+B(x-1)$ . Substituting the roots, $x=1$ gives $8=-A$ so $A=-8$ , and $x=2$ gives $11=B$ . Hence

\frac{3x+5}{x^2-3x+2} = \frac{-8}{x-1}+\frac{11}{x-2}.

Tip

With distinct linear factors, substitute each factor's root to knock out all but one term — much faster than equating coefficients. …

Part (b)Concept understanding — Integration By Completing Square

Integration by Completing the Square

You can integrate $\dfrac{1}{x^2+1}$ or $\dfrac{1}{x^2-a^2}$ on sight. But a quadratic denominator such as $x^2+4x+5$ fits neither standard form directly. The fix is to rewrite the quadratic as a perfect square plus (or minus) a constant, turning it into a form you already know.

The move

For $x^2+bx+c$ , add and subtract $\left(\dfrac{b}{2}\right)^2$ :

x^2+bx+c=\left(x+\frac{b}{2}\right)^2+\left(c-\frac{b^2}{4}\right).

After substituting $u=x+\dfrac{b}{2}$ the integral collapses to $\displaystyle\int\frac{du}{u^2+k^2}$ or $\displaystyle\int\frac{du}{u^2-k^2}$ .

Case A — leads to inverse tangent

\int\frac{dx}{x^2+4x+5}.

Complete the square: $x^2+4x+5=(x+2)^2+1$ . With $u=x+2$ ,

\int\frac{du}{u^2+1}=\tan^{-1}u+C=\tan^{-1}(x+2)+C.

Tip

$\displaystyle\int\frac{du}{u^2+a^2}=\frac1a\tan^{-1}\frac{u}{a}+C$ is the workhorse when the constant left over is positive.

Case B — leads to a logarithm

\int\frac{dx}{x^2-6x+5}.

Here $x^2-6x+5=(x-3)^2-4$ . With $u=x-3$ the denominator $u^2-4$ factors, so use partial fractions:

$\int\frac{du}{u^2-4}=\frac14\log\left|\frac{u-2}{u+2}\right|+C=\frac14\log\left|\frac{x-5}{x-1}\right|+C.$ …

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(a) Substitute $u=\sin x$ then split by partial fractions: $\frac12\ln\left|\frac{2+\sin x}{4+\sin x}\right|+C$ . (b) Split the numerator around the derivative of the denominator and complete the square: $\frac12\ln(x^2+4x+5)+\tan^{-1}(x+2)+C$ .

Part (a) — Substitution and partial fractions

The presence of $\cos x$ (the derivative of $\sin x$ ) in the numerator suggests $u=\sin x$ , so $du=\cos x\,dx$ and

\int\frac{\cos x}{(2+\sin x)(4+\sin x)}\,dx=\int\frac{du}{(2+u)(4+u)}.

Decompose $\dfrac1{(2+u)(4+u)}=\dfrac{A}{2+u}+\dfrac{B}{4+u}$ , i.e. $1=A(4+u)+B(2+u)$ .

$u=-2:\ 1=2A\Rightarrow A=\tfrac12$ .
$u=-4:\ 1=-2B\Rightarrow B=-\tfrac12$ .

Hence

$\int\left(\frac{1/2}{2+u}-\frac{1/2}{4+u}\right)du=\frac12\big(\ln|2+u|-\ln|4+u|\big)+C=\frac12\ln\left|\frac{2+u}{4+u}\right|+C.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 14 on this concept.

CBSE 2026Set A1 markMCQ
$\int\frac{dx}{x^2 - a^2} =$
(a) $\frac{1}{a}\tan^{-1}\frac{x}{a} + k$
(b) $\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + k$
(c) $\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right| + k$
(d) $\frac{1}{a}\log\left|\frac{x-a}{x+a}\right| + k$
›Reveal solutionSolution

Standard integral: $\int\tfrac{dx}{x^2-a^2}=\tfrac{1}{2a}\log\left|\tfrac{x-a}{x+a}\right|+k$ .

Using partial fractions, $\dfrac{1}{x^2-a^2}=\dfrac{1}{(x-a)(x+a)}=\dfrac{1}{2a}\left(\dfrac{1}{x-a}-\dfrac{1}{x+a}\right)$ .

…

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CBSE 2026Set ANNUAL1 mark
$\int \frac{dx}{\sqrt{2x - x^2}}$
›Reveal solutionSolution

Complete the square under the root, then use the standard integral $\int dx/\sqrt{a^2-x^2}=\sin^{-1}(x/a)+c$ .

$2x-x^2 = 1-(x-1)^2$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. \int dx/(x² - 2x + 2) is ................. .$
(a) tan⁻¹(x−1) + c
(b) tan⁻¹(x+1) + c
(c) tan⁻¹(x+2) + c
(d) tan⁻¹(x−2) + c
›Reveal solutionSolution

Complete the square in the denominator, then use the standard $\int \frac{dt}{t^2+a^2}$ form.

$x^2-2x+2 = (x-1)^2+1$

…

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CBSE 2025Set ANNUAL1 mark
$\displaystyle\int\sqrt{x^2+3x+\dfrac{9}{4}}\,dx=$
›Reveal solutionSolution

The expression under the root is a perfect square, so the square root simplifies to a linear expression.

Note that $x^2+3x+\dfrac{9}{4}=\left(x+\dfrac{3}{2}\right)^2$ .

So $\sqrt{x^2+3x+\dfrac{9}{4}}=\left|x+\dfrac{3}{2}\right|$ , which (taking the positive branch) is $x+\dfrac{3}{2}$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$\displaystyle\int \sqrt{1+x^2}\, dx$
(a) $\dfrac{x}{2}\sqrt{1+x^2} + \dfrac{1}{2}\log\left|x+\sqrt{1+x^2}\right| + c$
(b) $\dfrac{2}{3}(1+x^2)^{3/2} + c$
(c) $\dfrac{2}{3}x(1+x^2)^{3/2} + c$
(d) $\dfrac{x^2}{2}\sqrt{1+x^2} + \dfrac{1}{2}x^2\log\left|x+\sqrt{1+x^2}\right| + c$
›Reveal solutionSolution

This is a standard integral of the form $\int\sqrt{x^2+a^2}\,dx$ .

The standard formula (derivable by integration by parts, treating $\sqrt{1+x^2}=\sqrt{1+x^2}\cdot 1$ ) is:

$\int \sqrt{x^2+a^2}\,dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$

With $a=1$ : …

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CBSE 2024Set D1 markMCQ
$\int \frac{dx}{x(x+2)} =$
(a) $\log\left|\frac{x}{x+2}\right| + c$
(b) $\frac{1}{2}\log\left|\frac{x}{x+2}\right| + c$
(c) $\log|x| + c$
(d) $\log|x+2| + c$
›Reveal solutionSolution

Partial fractions give $\frac{1}{2}\log\left|\frac{x}{x+2}\right|+c$ .

Write $\dfrac{1}{x(x+2)} = \dfrac{1}{2}\left(\dfrac{1}{x} - \dfrac{1}{x+2}\right)$ .

…

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CBSE 2024Set D1 markMCQ
$\int \sqrt{a^2 - x^2}\,dx =$
(a) $\frac{x}{2}\sqrt{a^2 - x^2}\,dx$
(b) $\frac{a^2}{2}\sin^{-1}\frac{x}{a} + c$
(c) $\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + c$
(d) $\frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\sin^{-1}\frac{x}{a} + c$
›Reveal solutionSolution

Standard result: $\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+c$ .

This is a memorised standard form (derivable by the substitution $x=a\sin\theta$ ):

…

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CBSE 2023Set E1 markMCQ
$\displaystyle\int\dfrac{3\,dx}{\sqrt{1-9x^{2}}} =$
(a) $\tan^{-1}3x+k$
(b) $\sec^{-1}3x+k$
(c) $\sin^{-1}3x+k$
(d) $\cos^{-1}3x+k$
›Reveal solutionSolution

With $u=3x$ , $du=3\,dx$ , the integral becomes $\int\dfrac{du}{\sqrt{1-u^{2}}}=\sin^{-1}3x+k$ .

Let $u=3x$ , so $du=3\,dx$ . The numerator $3\,dx=du$ .

…

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CBSE 2022Set ANNUAL1 markMCQ
$\int \frac{dx}{x^2 - 1} =$
(a) $\sin^{-1} x + k$
(b) $\frac{1}{2}\log\left|\frac{x-1}{x+1}\right| + k$
(c) $\frac{1}{2}\log\left|\frac{x+1}{x-1}\right| + k$
(d) $\sqrt{1 - x^2} + k$
›Reveal solutionSolution

$\int \frac{dx}{x^2-1} = \frac{1}{2}\log\left|\frac{x-1}{x+1}\right| + k$ .

The standard result is $\int \frac{dx}{x^2-a^2} = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right| + k$ .

…

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CBSE 2021Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{x\,dx}{(x-1)(x-2)}$
(a) $\log\left|\dfrac{(x-1)^{2}}{x-2}\right|+C$
(b) $\log\left|\dfrac{(x-2)^{2}}{x-1}\right|+C$
(c) $\log\left|\left(\dfrac{x-1}{x-2}\right)^{2}\right|+C$
(d) $\log\left|(x-1)(x-2)\right|+C$
›Reveal solutionSolution

Partial fractions give $\dfrac{x}{(x-1)(x-2)} = \dfrac{-1}{x-1}+\dfrac{2}{x-2}$ , integrating to $\log\left|\dfrac{(x-2)^2}{x-1}\right|+C$ .

Let $\dfrac{x}{(x-1)(x-2)} = \dfrac{A}{x-1}+\dfrac{B}{x-2}$

$x = A(x-2)+B(x-1)$

At $x=1$ : $1 = -A \Rightarrow A=-1$

At $x=2$ : $2 = B \Rightarrow B=2$

…

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CBSE 2021Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{dx}{x^{2}+2x+2}$
(a) $x\tan^{-1}(x+1)+C$
(b) $\tan^{-1}(x+1)+C$
(c) $(x+1)\tan^{-1}x+C$
(d) $\tan^{-1}x+C$
›Reveal solutionSolution

Completing the square, $x^2+2x+2=(x+1)^2+1$ , giving $\tan^{-1}(x+1)+C$ .

$x^2+2x+2 = (x+1)^2+1$

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. If 1/(x(x-3)) = A/x + B/(x-3), then the value of B is:$
(a) 1/2
(b) −1/3
(c) 1/3
(d) −1/2
›Reveal solutionSolution

Clear the denominator and compare coefficients (or plug in $x=3$ ) to isolate $B$ .

Write $\dfrac{1}{x(x-3)} = \dfrac{A}{x} + \dfrac{B}{x-3}$ .

Multiplying both sides by $x(x-3)$ :

$1 = A(x-3) + Bx$

…

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Q.(a) Find: $\int \frac{\cos x}{(2 + \sin x)(4 + \sin x)}\, dx$

Part (a)Concept understanding — Partial Fraction Decomposition

Partial Fraction Decomposition

When it applies

The standard forms

Worked example

Part (b)Concept understanding — Integration By Completing Square

Integration by Completing the Square

The move

Case A — leads to inverse tangent

Case B — leads to a logarithm

Part (a) — $\displaystyle\int\frac{\cos x}{(2+\sin x)(4+\sin x)}\,dx$

Part (a) — Substitution and partial fractions

Q.(a) Find: ∫cos⁡x(2+sin⁡x)(4+sin⁡x) dx\int \frac{\cos x}{(2 + \sin x)(4 + \sin x)}\, dx∫(2+sinx)(4+sinx)cosx​dx

Part (a) — Substitution and partial fractions

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Q.(a) Find: $\int \frac{\cos x}{(2 + \sin x)(4 + \sin x)}\, dx$

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