JEE Main 2018 (15 April Evening)

LEVELJEE Main

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{2 x + 5}{7 - 6 x - x ^{2}} d x = A 7 - 6 x - x^{2} + B sin^{- 1} (\frac{x + 3}{4}) + C$ (Where C is a constant of integration), then the ordered pair (A, B) is equal to :-

Select Answer:

Visualized Solution

$\int \frac{p x + q}{a x ^{2} + b x + c} d x$

The integral is of the form: $\int \frac{p x + q}{a x ^{2} + b x + c} d x$
Numerator: $2 x + 5$
Quadratic in denominator: $7 - 6 x - x^{2}$

$N u m er a t or = k_{1} \frac{d}{d x} (D e n o mina t or) + k_{2}$

Strategy: Express $2 x + 5 = k_{1} \frac{d}{d x} (7 - 6 x - x^{2}) + k_{2}$
This allows us to split the integral into a substitution part and a standard formula part.

$\frac{d}{d x} (7 - 6 x - x^{2})$

Let's find the derivative of the quadratic term.
$\frac{d}{d x} (7 - 6 x - x^{2}) = 0 - 6 - 2 x = - 2 x - 6$

$2 x + 5 = k_{1} (- 2 x - 6) + k_{2}$

Substitute the derivative back into our strategy equation.
$2 x + 5 = k_{1} (- 2 x - 6) + k_{2}$
Expand the right side: $2 x + 5 = - 2 k_{1} x - 6 k_{1} + k_{2}$

Comparing coefficients of $x$

Comparing coefficients of $x$ on both sides:
$2 = - 2 k_{1}$
$⟹ k_{1} = - 1$

Comparing constant terms

Comparing constant terms on both sides:
$5 = - 6 k_{1} + k_{2}$
Substitute $k_{1} = - 1$ : $5 = - 6 (- 1) + k_{2}$
$5 = 6 + k_{2} ⟹ k_{2} = - 1$

$I = I_{1} + I_{2}$

Rewrite the original integral using $k_{1}$ and $k_{2}$ :
$I = \int \frac{- 1 ( - 2 x - 6 ) - 1}{7 - 6 x - x ^{2}} d x$
Split into two separate integrals:
$I = - \int \frac{- 2 x - 6}{7 - 6 x - x ^{2}} d x - \int \frac{1}{7 - 6 x - x ^{2}} d x$

Solving $I_{1}$ via Substitution

$I_{1} = - \int \frac{- 2 x - 6}{7 - 6 x - x ^{2}} d x$
Let $t = 7 - 6 x - x^{2} ⟹ d t = (- 2 x - 6) d x$
$I_{1} = - \int \frac{1}{t} d t = - \int t^{- \frac{1}{2}} d t$

Evaluating $I_{1}$ and finding $A$

$I_{1} = - \frac{t ^{\frac{1}{2}}}{\frac{1}{2}} = - 2 t$
Substitute back $t$ : $I_{1} = - 2 7 - 6 x - x^{2}$
Comparing with $A 7 - 6 x - x^{2}$ , we get $A = - 2$

Preparing $I_{2}$

$I_{2} = - \int \frac{1}{7 - 6 x - x ^{2}} d x$
Goal: Complete the square for the quadratic expression $7 - 6 x - x^{2}$ .

Completing the Square

$7 - 6 x - x^{2} = - (x^{2} + 6 x - 7)$
Add and subtract $(\frac{6}{2})^{2} = 9$ inside the bracket:
$= - (x^{2} + 6 x + 9 - 9 - 7)$
$= - ((x + 3)^{2} - 16)$
$= 16 - (x + 3)^{2} = 4^{2} - (x + 3)^{2}$

Standard Integral Formula

The integral becomes: $I_{2} = - \int \frac{d x}{4 ^{2} - ( x + 3 ) ^{2}}$
Standard Formula: $\int \frac{d x}{a ^{2} - x ^{2}} = sin^{- 1} (\frac{x}{a}) + C$
Here $a = 4$ and $x$ is replaced by $(x + 3)$ .

Evaluating $I_{2}$ and finding $B$

Applying the formula: $I_{2} = - sin^{- 1} (\frac{x + 3}{4})$
The problem states this part is $B sin^{- 1} (\frac{x + 3}{4})$
Comparing the two expressions, we get $B = - 1$

Final Conclusion

Total Integral $I = - 2 7 - 6 x - x^{2} - sin^{- 1} (\frac{x + 3}{4}) + C$
Given form: $A 7 - 6 x - x^{2} + B sin^{- 1} (\frac{x + 3}{4}) + C$
We found $A = - 2$ and $B = - 1$ .
Ordered pair $(A, B) = (- 2, - 1)$

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The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The integral

\int \frac{2 x + 5}{7 - 6 x - x ^{2}} d x

presents a classic JEE Advanced pattern: a linear polynomial divided by the square root of a quadratic.

We cannot use simple substitution immediately because the numerator is not the derivative of the quadratic expression under the radical. Instead, we must employ Decomposition Mode.

The Art of Decomposition

Our strategy is to express the numerator

2 x + 5

as a linear combination of the derivative of the quadratic

f (x) = 7 - 6 x - x^{2}

and a constant.

The derivative is

f^{'} (x) = - 2 x - 6

. We set up the equation:

2 x + 5 = k_{1} (- 2 x - 6) + k_{2}

By splitting the integral into two parts,

I_{1} = \int \frac{k _{1} ( - 2 x - 6 )}{f ( x )} d x

and

I_{2} = \int \frac{k _{2}}{f ( x )} d x

, we can solve the problem systematically.

I_{1}

is solved via substitution, and

I_{2}

is solved by completing the square.

The First Victory

Solving

I_{1}

Comparing coefficients in

2 x + 5 = - 2 k_{1} x - 6 k_{1} + k_{2}

: 1. For

x

2 = - 2 k_{1} \Rightarrow k_{1} = - 1

2. For constants:

5 = - 6 k_{1} + k_{2} \Rightarrow 5 = 6 + k_{2} \Rightarrow k_{2} = - 1

The integral becomes:

I = - \int \frac{- 2 x - 6}{7 - 6 x - x ^{2}} d x - \int \frac{1}{7 - 6 x - x ^{2}} d x

For

I_{1}

, let

t = 7 - 6 x - x^{2}

, so

d t = (- 2 x - 6) d x

. The integral evaluates to:

- \int t^{- 1/2} d t = - 2 t = - 2 7 - 6 x - x^{2}

Thus, we identify

A = - 2

The Second Victory

Completing the Square for

I_{2}

For

I_{2} = - \int \frac{d x}{7 - 6 x - x ^{2}}

, we complete the square for the quadratic:

7 - 6 x - x^{2} = - (x^{2} + 6 x - 7) = - (x^{2} + 6 x + 9 - 16) = 16 - (x + 3)^{2}

Substituting this back into the integral:

I_{2} = - \int \frac{d x}{4 ^{2} - ( x + 3 ) ^{2}}

Using the standard form

\int \frac{d x}{a ^{2} - x ^{2}} = sin^{- 1} (\frac{x}{a})

, we get:

I_{2} = - sin^{- 1} (\frac{x + 3}{4})

Comparing this to

B sin^{- 1} (\frac{x + 3}{4})

, we find

B = - 1

The Grand Finale

Combining both parts, the total integral is:

I = - 2 7 - 6 x - x^{2} - sin^{- 1} (\frac{x + 3}{4}) + C

The resulting ordered pair is $(A, B) = (- 2, - 1)$ . By focusing on structure and decomposition, we have successfully dismantled the integral.

Common Traps & FAQs

Wait, why can't we just use substitution

u = 7 - 6 x - x^{2}

right away?

That is a fantastic question! Substitution works perfectly when the numerator is exactly the derivative of the denominator (or a constant multiple of it). Here, the derivative of

7 - 6 x - x^{2}

- 2 x - 6

. Our numerator is

2 x + 5

. They are close, but not identical. By decomposing the numerator into a part that matches the derivative and a constant part, we create two integrals: one that is solvable by substitution and one that is a standard inverse trigonometric form.

How do I know when to complete the square?

Whenever you encounter a quadratic expression under a square root in the denominator, completing the square is your most powerful tool. It transforms the quadratic into a standard form like

a^{2} - x^{2}

, which is the gateway to the

sin^{- 1}

integral formula. It is a systematic, reliable process that turns chaos into order.

Why did we get a negative sign for

k_{1}

Don't worry about the negative sign; it is just the math correcting itself! The coefficient of

x

in our numerator is

2

, but the derivative of the quadratic is

- 2 x

. To make them match, we need a factor of

- 1

. It is not a mistake; it is simply the necessary adjustment to align our numerator with the derivative of the denominator.

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{2 x + 5}{7 - 6 x - x ^{2}} d x = A 7 - 6 x - x^{2} + B sin^{- 1} (\frac{x + 3}{4}) + C$ (Where C is a constant of integration), then the ordered pair (A, B) is equal to :-

Select Answer:

Visualized Solution

$\int \frac{p x + q}{a x ^{2} + b x + c} d x$

$N u m er a t or = k_{1} \frac{d}{d x} (D e n o mina t or) + k_{2}$

$\frac{d}{d x} (7 - 6 x - x^{2})$

$2 x + 5 = k_{1} (- 2 x - 6) + k_{2}$

Comparing coefficients of $x$

Comparing constant terms

$I = I_{1} + I_{2}$

Solving $I_{1}$ via Substitution

Evaluating $I_{1}$ and finding $A$

Preparing $I_{2}$

Completing the Square

Standard Integral Formula

Evaluating $I_{2}$ and finding $B$

Final Conclusion

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The Art of Decomposition

The First Victory

The Second Victory

The Grand Finale

Similar Questions

If $\int \frac{t a n x}{1 + t a n x + t a n ^{2} x} d x = x - \frac{K}{A} tan^{- 1} (\frac{K t a n x + 1}{A}) + C$ , (C is a constant of integration), then the ordered pair (K, A) is equal to :

Let $I_{n} = \int tan^{n} x d x, (n > 1)$ . If $I_{4} + I_{6} = a tan^{5} x + b x^{5} + C$ , where C is a constant of integration, then the ordered pair $(a, b)$ is equal to:

If $\int \frac{c o s θ}{5 + 7 s i n θ - 2 c o s ^{2} θ} d θ = A lo g_{e} ∣ B (θ) ∣ + C$ where $C$ is a constant of integration, then $\frac{B ( θ )}{A}$ can be:

If $\int e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}} + \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}) d x = g (x) + C,$ where C is the constant of integration, then $g (\frac{1}{2})$ equals:

If $\int \frac{5 t a n x}{t a n x - 2} d x = x + a ln ∣ sin x - 2 cos x ∣ + k$ , then $a$ is equal to

If $\int \frac{d x}{( x ^{2} - 2 x + 10 ) ^{2}} = A (tan^{- 1} (\frac{x - 1}{3}) + \frac{f ( x )}{x ^{2} - 2 x + 10}) + C$ , where $C$ is a constant of integration, then :

The integral $\int ((\frac{x}{2})^{x} + (\frac{2}{x})^{x}) lo g_{2} x d x$ is equal to

The integral $\int \frac{( 1 - \frac{1}{3} ) ( c o s x - s i n x )}{1 + \frac{2}{3} s i n 2 x} d x$ is equal to

If $\int \frac{2 x ^{2} + 5 x + 9}{x ^{2} + x + 1} d x = x x^{2} + x + 1 + α x^{2} + x + 1 + β lo g_{e} ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣ + C .$ where C is the constant of integration, then $α + 2 β$ is equal to

$\int {\frac{l o g x - 1}{1 + ( l o g x ) ^{2}}}^{2} d x$ is equal to

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫ax2+bx+c​px+q​dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Numerator=k1​dxd​(Denominator)+k2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxd​(7−6x−x2)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2x+5=k1​(−2x−6)+k2​

Comparing coefficients of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Comparing constant terms

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Solving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​ via Substitution

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I1​ and finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Preparing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​

Completing the Square

Standard Integral Formula

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​ and finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Final Conclusion

The Sigma Insight: Evaluation of Special Integral Forms

Analyzing the Setup

The Art of Decomposition

The First Victory

The Second Victory

The Grand Finale

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let In​=∫tannxdx,(n>1). If I4​+I6​=atan5x+bx5+C, where C is a constant of integration, then the ordered pair (a,b) is equal to:

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫x2+x+1​2x2+5x+9​dx=xx2+x+1​+αx2+x+1​+β loge​∣x+21​+x2+x+1​∣+C. where C is the constant of integration, then α+2β is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫{1+(logx)2logx−1​}2dx is equal to

$\int \frac{p x + q}{a x ^{2} + b x + c} d x$

$N u m er a t or = k_{1} \frac{d}{d x} (D e n o mina t or) + k_{2}$

$\frac{d}{d x} (7 - 6 x - x^{2})$

$2 x + 5 = k_{1} (- 2 x - 6) + k_{2}$

Comparing coefficients of $x$

$I = I_{1} + I_{2}$

Solving $I_{1}$ via Substitution

Evaluating $I_{1}$ and finding $A$

Preparing $I_{2}$

Evaluating $I_{2}$ and finding $B$

If $\int \frac{t a n x}{1 + t a n x + t a n ^{2} x} d x = x - \frac{K}{A} tan^{- 1} (\frac{K t a n x + 1}{A}) + C$ , (C is a constant of integration), then the ordered pair (K, A) is equal to :

Let $I_{n} = \int tan^{n} x d x, (n > 1)$ . If $I_{4} + I_{6} = a tan^{5} x + b x^{5} + C$ , where C is a constant of integration, then the ordered pair $(a, b)$ is equal to:

If $\int \frac{c o s θ}{5 + 7 s i n θ - 2 c o s ^{2} θ} d θ = A lo g_{e} ∣ B (θ) ∣ + C$ where $C$ is a constant of integration, then $\frac{B ( θ )}{A}$ can be:

If $\int e^{x} (\frac{x s i n ^{- 1} x}{1 - x ^{2}} + \frac{s i n ^{- 1} x}{( 1 - x ^{2} ) ^{3/2}} + \frac{x}{1 - x ^{2}}) d x = g (x) + C,$ where C is the constant of integration, then $g (\frac{1}{2})$ equals:

If $\int \frac{5 t a n x}{t a n x - 2} d x = x + a ln ∣ sin x - 2 cos x ∣ + k$ , then $a$ is equal to

If $\int \frac{d x}{( x ^{2} - 2 x + 10 ) ^{2}} = A (tan^{- 1} (\frac{x - 1}{3}) + \frac{f ( x )}{x ^{2} - 2 x + 10}) + C$ , where $C$ is a constant of integration, then :

The integral $\int ((\frac{x}{2})^{x} + (\frac{2}{x})^{x}) lo g_{2} x d x$ is equal to

The integral $\int \frac{( 1 - \frac{1}{3} ) ( c o s x - s i n x )}{1 + \frac{2}{3} s i n 2 x} d x$ is equal to

If $\int \frac{2 x ^{2} + 5 x + 9}{x ^{2} + x + 1} d x = x x^{2} + x + 1 + α x^{2} + x + 1 + β lo g_{e} ∣ x + \frac{1}{2} + x^{2} + x + 1 ∣ + C .$ where C is the constant of integration, then $α + 2 β$ is equal to

$\int {\frac{l o g x - 1}{1 + ( l o g x ) ^{2}}}^{2} d x$ is equal to