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TG EAMCET 3 May 2025 Shift 2 Paper

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Question : 72 of 160

Marks: +1, -0

\int \frac{3^x(x \log 3-1)}{x^2} dx=

$x \cdot 3^x+c$
$\frac{3^x}{x^2}+c$
$x^2 3^x+c$
$\frac{3^x}{x}+c$

Solution:

Let's call the integral

I

.

I=\int \frac{3^x(x \log 3-1)}{x^2} dx

Let's make a substitution. Set

t=\frac{3^x}{x}

.

Now, calculate the derivative of

t

with respect to

x

:

\frac{dt}{dx}=\frac{3^x \log 3 \cdot x-3^x}{x^2}=\frac{3^x(x \log 3-1)}{x^2}

So,

dt=\frac{3^x(x \log 3-1)}{x^2} dx

.

This shows that the part inside our integral,

\frac{3^x(x \log 3-1)}{x^2} dx

, is simply

dt

.

So the integral becomes

I=\int dt=t+C

.

Recall

t=\frac{3^x}{x}

, so the answer is

I=\frac{3^x}{x}+C

.

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