Let I(x) = x+7x dx and I(9) = 12 + 7 e 7. If I(1) = + 7 e (1+2 2), then ^4 is equal to:

Given: I(x) = x+7x dx and I(9) = 12 + 7 7. Find: ^4 if I(1) = + 7 (1+2 2). Use the substitution shown in the solution: x = t^2, dx = 2t dt Then x+7x dx = 2 t^2+7t^2 t dt = 2 t^2+7 dt From the extracted working, I(t) = 2 [ t2 t^2+7 + 72 | t+ t^2+7 | ] + C So, I(t) = t t^2+7 + 7 | t+ t^2+7 | + C Substituting back t= x, I(x) = x x+7 + 7 | x+ x+7 | + C Now use I(9) = 12 + 7 7: I(9) = 9 16 + 7 | 9+ 16 | + C 12 + 7 7 = 12 + 7 (3+4) + C Hence, C=0 Therefore, I(x) = x x+7 + 7 | x+ x+7 | Now substitute x=1: I(1) = 1 8 + 7 | 1+ 8 | I(1) = 8 + 7 (1+ 8) Since 8 = 2 2, this becomes I(1) = 2 2 + 7 (1+2 2) Comparing with I(1) = + 7 (1+2 2), we get = 2 2 = 8 Hence, ^4 = ( 8)^4 = 8^2 = 64 Therefore, the correct option is A. After finding I(1) = 8 + 7 (1+ 8) use 8 = 2 2 to rewrite it as I(1) = 2 2 + 7 (1+2 2) Now compare term-by-term with the given form I(1) = + 7 (1+2 2) So the logarithmic parts are identical, which gives = 2 2 Then ^2 = 8 ^4 = 64

Let I(x) = x+7x dx and I(9) = 12 + 7 e 7. If I(1) = + 7 e (1+2 2)… — Solution | JEE Mathematics

MCQMediumJEE 2023Indefinite Integrals

Let $I(x)$ = $\int \sqrt{\frac{x+7}{x}} \, dx$ and $I(9) = 12 + 7 \log_e 7$ . If $I(1) = \alpha + 7 \log_e (1+2\sqrt{2})$ , then $\alpha^4$ is equal to:

A
$64$
B
$128$
C
$32$
D
$16$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $I(x) = \int \sqrt{\frac{x+7}{x}} \, dx$ and $I(9) = 12 + 7 \ln 7$ .

Find: $\alpha^4$ if $I(1) = \alpha + 7 \ln(1+2\sqrt{2})$ .

Use the substitution shown in the solution:

x = t^2, \quad dx = 2t \, dt

Then

\int \sqrt{\frac{x+7}{x}} \, dx = 2 \int \sqrt{\frac{t^2+7}{t^2}} \, t \, dt = 2 \int \sqrt{t^2+7} \, dt

From the extracted working,

I(t) = 2 \left[ \frac{t}{2}\sqrt{t^2+7} + \frac{7}{2}\ln \left| t+\sqrt{t^2+7} \right| \right] + C

So,

I(t) = t\sqrt{t^2+7} + 7\ln \left| t+\sqrt{t^2+7} \right| + C

Substituting back $t=\sqrt{x}$ ,

I(x) = \sqrt{x}\sqrt{x+7} + 7\ln \left| \sqrt{x}+\sqrt{x+7} \right| + C

Now use $I(9) = 12 + 7\ln 7$ :

I(9) = \sqrt{9}\sqrt{16} + 7\ln \left| \sqrt{9}+\sqrt{16} \right| + C

12 + 7\ln 7 = 12 + 7\ln(3+4) + C

Hence,

C=0

Therefore,

I(x) = \sqrt{x}\sqrt{x+7} + 7\ln \left| \sqrt{x}+\sqrt{x+7} \right|

Now substitute $x=1$ :

I(1) = \sqrt{1}\sqrt{8} + 7\ln \left| \sqrt{1}+\sqrt{8} \right|

I(1) = \sqrt{8} + 7\ln(1+\sqrt{8})

Since $\sqrt{8} = 2\sqrt{2}$ , this becomes

I(1) = 2\sqrt{2} + 7\ln(1+2\sqrt{2})

Comparing with $I(1) = \alpha + 7\ln(1+2\sqrt{2})$ , we get

\alpha = 2\sqrt{2} = \sqrt{8}

Hence,

\alpha^4 = (\sqrt{8})^4 = 8^2 = 64

Therefore, the correct option is A.

Comparison Step for \alpha

After finding

I(1) = \sqrt{8} + 7\ln(1+\sqrt{8})

use $\sqrt{8} = 2\sqrt{2}$ to rewrite it as

I(1) = 2\sqrt{2} + 7\ln(1+2\sqrt{2})

Now compare term-by-term with the given form

I(1) = \alpha + 7\ln(1+2\sqrt{2})

So the logarithmic parts are identical, which gives

\alpha = 2\sqrt{2}

Then

\alpha^2 = 8 \quad \Rightarrow \quad \alpha^4 = 64

Common mistakes

Taking $\alpha = \sqrt{8}$ but not simplifying $\ln(1+\sqrt{8})$ to $\ln(1+2\sqrt{2})$ . This hides the direct comparison with the given expression. First rewrite $\sqrt{8}=2\sqrt{2}$ , then match terms carefully.
Forgetting to use the condition $I(9)=12+7\ln 7$ to determine the constant of integration. An indefinite integral always includes a constant. Substitute $x=9$ before evaluating $I(1)$ .
Computing $\alpha^4$ incorrectly from $\alpha=2\sqrt{2}$ . Since $\alpha^2=8$ , we get $\alpha^4=(\alpha^2)^2=64$ . Do not stop at $8$ or square only the coefficient.

Practice more Indefinite Integrals questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Indefinite Integrals

Answer

Step-by-step solution

Common mistakes

Practice more Indefinite Integrals questions

Related questions