Indefinite IntegrationmediumFree

Sum of f+f²+… from ∫eˣ((x+2)/(x+4))² dx | JEE

JEE Maths question with a full step-by-step solution.

Question

\displaystyle\int e^{x}\left(\dfrac{x+2}{x+4}\right)^{2}dx=\dfrac{f(x)\,e^{x}}{x+4}+C

, then

f(x)+f^{2}(x)+f^{3}(x)+\cdots\infty

x=\dfrac{1}{2}

\dfrac{1}{2}

\dfrac{1}{4}

1

correct

\dfrac{3}{2}

Solution

Step 1: Find

f(x)

via the

e^{x}(g+g')

pattern. Take

g(x)=\dfrac{x}{x+4}

, so

g'(x)=\dfrac{(x+4)-x}{(x+4)^{2}}=\dfrac{4}{(x+4)^{2}}

Step 2: Add

g+g'

over

(x+4)^{2}

. Writing

g=\dfrac{x(x+4)}{(x+4)^{2}}

g+g'=\dfrac{x(x+4)+4}{(x+4)^{2}}=\dfrac{x^{2}+4x+4}{(x+4)^{2}}=\dfrac{(x+2)^{2}}{(x+4)^{2}}=\left(\dfrac{x+2}{x+4}\right)^{2}

which is the bracket in the integrand. Step 3: Apply

\displaystyle\int e^{x}(g+g')\,dx=e^{x}g

\int e^{x}\left(\dfrac{x+2}{x+4}\right)^{2}dx=e^{x}\cdot\dfrac{x}{x+4}+C

Comparing with

\dfrac{f(x)e^{x}}{x+4}+C

gives

f(x)=x

. Step 4: Sum the geometric series

f+f^{2}+f^{3}+\cdots=\dfrac{f}{1-f}

for

|f|<1

. At

x=\dfrac{1}{2}

f=\dfrac{1}{2}

\dfrac{1/2}{1-1/2}=1

Correct answer: (3)

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