(a) Evaluate : $\int \frac{3x^2 + 6x + 1}{(x + 3)(x^2 + 1)}\, dx$. OR (b) Find an initial basic feasible solution of following problem using north west corner rule. | | $D_1$ | $D_2$ | $D_3$ | $D_4$ | Supply | | --- | --- | --- | --- | --- | --- | | $O_1$ | $5$ | $3$ | $6$ | $2$ | $19$ | | $O_2$ | $4$ | $7$ | $9$ | $1$ | $37$ | | $O_3$ | $3$ | $4$ | $7$ | $5$ | $34$ | | Demand | $16$ | $18$ | $31$ | $25$ | |

(a) Evaluate : ∫ (3x² + 6x + 1)/((x + 3)(x² + 1)) dx. OR (b) Find an initial basic feasible solution of follow

Question 22 of 42

Evaluate : $\int \frac{3x^2 + 6x + 1}{(x + 3)(x^2 + 1)}\, dx$ . OR
Find an initial basic feasible solution of following problem using north west corner rule.

	$D_1$	$D_2$	$D_3$	$D_4$	Supply
$O_1$	$5$	$3$	$6$	$2$	$19$
$O_2$	$4$	$7$	$9$	$1$	$37$
$O_3$	$3$	$4$	$7$	$5$	$34$
Demand	$16$	$18$	$31$	$25$

Tamil Nadu DgeTamil Nadu HSC (DGE) Commerce Board 2022Subjective· 5mImportance★★★★★

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(a) Partial fractions give $\frac{1}{x+3}+\frac{2x}{x^2+1}$ , so the integral is $\ln|x+3|+\ln(x^2+1)+C$ . (b) NW-corner allocations cost $580$ .

(a) Partial fractions.

Let $\dfrac{3x^2+6x+1}{(x+3)(x^2+1)}=\dfrac{A}{x+3}+\dfrac{Bx+C}{x^2+1}.$ Then

$3x^2+6x+1=A(x^2+1)+(Bx+C)(x+3).$

Put $x=-3$ : $27-18+1=10=A(10)\Rightarrow A=1.$ Compare $x^2$ : $3=A+B\Rightarrow B=2.$ Compare constants: $1=A+3C\Rightarrow C=0.$ (Check $x$ -term: $6=3B+C=6.$ )

Hence

$\int\frac{3x^2+6x+1}{(x+3)(x^2+1)}\,dx=\int\frac{1}{x+3}\,dx+\int\frac{2x}{x^2+1}\,dx=\ln|x+3|+\ln(x^2+1)+C.$

(b) North-west corner rule.

Total supply $=19+37+34=90=$ total demand $=16+18+31+25$ , so the problem is balanced. Allocating from the top-left cell:

	$D_1$	$D_2$	$D_3$	$D_4$	Supply
$O_1$	16	3			19
$O_2$		15	22		37
$O_3$			9	25	34

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

CBSE 2026Set ANNUAL1 mark
$\int \frac{x}{(1 + x)(2 + x)}\,dx = \int \left(\frac{A}{1 + x} + \frac{B}{2 + x}\right)dx$
›Reveal solutionSolution

Partial fractions give $A = -1,\; B = 2$ , so the claim $A = 1,\, B = 2$ is False.

Decompose the integrand:

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

Multiply through by $(1+x)(2+x)$ :

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

Put $x = -1$ : $\;-1 = A(2 - 1) + B(0) = A \Rightarrow A = -1.$

…

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CBSE 2024Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{dx}{(x - 8)(x + 7)} =$
(a) $\dfrac{1}{15} \log \left|\dfrac{x + 2}{x - 1}\right| + c$
(b) $\dfrac{1}{15} \log \left|\dfrac{x + 8}{x + 7}\right| + c$
(c) $\dfrac{1}{15} \log \left|\dfrac{x - 8}{x + 7}\right| + c$
(d) $(x - 8)(x - 7) + c$
(e) $\dfrac{1}{15} \log \left|\dfrac{x + 2}{x + 1}\right| + c$
(f) $(x - 8)(x + 7) + c$
›Reveal solutionSolution

Partial fractions give $\dfrac{1}{(x-8)(x+7)} = \dfrac{1}{15}\!\left(\dfrac{1}{x-8} - \dfrac{1}{x+7}\right)$ , which integrates to $\dfrac{1}{15}\log\left|\dfrac{x-8}{x+7}\right| + c$ .

Write

$\frac{1}{(x-8)(x+7)} = \frac{A}{x-8} + \frac{B}{x+7} \;\Rightarrow\; A(x+7) + B(x-8) = 1.$

Put $x = 8$ : $\;A(15) = 1 \Rightarrow A = \dfrac{1}{15}$ .

Put $x = -7$ : $\;B(-15) = 1 \Rightarrow B = -\dfrac{1}{15}$ .

Hence …

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CBSE 2022Set ANNUAL1 mark
$\int \dfrac{x}{(x + 2)(x + 3)}\, dx =$
›Reveal solutionSolution
Partial fractions give $\dfrac{x}{(x+2)(x+3)} = \dfrac{-2}{x+2} + \dfrac{3}{x+3}$ , so the blank preceding $\displaystyle\int \dfrac{3}{x+3}\,dx$ is $\displaystyle\int \dfrac{-2}{x+2}\,dx$ .

Resolve the integrand into partial fractions:

$\frac{x}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}.$

Clearing denominators: $x = A(x+3) + B(x+2)$ .

Put $x = -2$ : $-2 = A(1) \Rightarrow A = -2$ .

Put $x = -3$ : $-3 = B(-1) \Rightarrow B = 3$ .

Hence

$\frac{x}{(x+2)(x+3)} = \frac{-2}{x+2} + \frac{3}{x+3},$

so …
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