≤ft({x-3}{x²+9})² d x=

Correct Answer is : 13tan⁡−1(x3)+3x2+9+c1/3 ^{-1}≤ft(x/3)+{3}{x²+9}+c31tan−1(3x)+x2+93+c, where ccc is the constant of integration.

1/3 ^{-1}≤ft(x/3)-{3}{x²+9}+c, where c is the constant of integration.

1/3 ^{-1}≤ft(x/3)-{1}{x²+9}+c, where c is the constant of integration.

1/3 ^{-1}≤ft(x/3)+{3}{x²+9}+c, where c is the constant of integration.

1/3 ^{-1}≤ft(x/3)-{1}{x²+9}+c, where c is the constant of integration. Let {aligned} I & =≤ft({x-3}{x²+9})² d x \\ & = {x²-6 x+9}{≤ft(x²+9)²} d x \\ & =≤ft[{1}{x²+9}-{6 x}{≤ft(x²+9)²}] d x \\ & = {1}{x²+3²} d x-3 {2 x}{≤ft(x²+9)²} d x {aligned} In 2^{ {nd }} integral, Put x²+9={t} 2 x d x={dt} {aligned} ∴ {I} & =1/3 ^{-1}≤ft(x/3)-3 {{dt}}{{t}²} \\ & =1/3 ^{-1}≤ft(x/3)-3≤ft({-1}{{t}})+c \\ & =1/3 ^{-1}≤ft(x/3)+{3}{x²+9}+c {aligned}

≤ft({x-3}{x²+9})² d x=

Correct Answer is : 13tan⁡−1(x3)+3x2+9+c1/3 ^{-1}≤ft(x/3)+{3}{x²+9}+c31tan−1(3x)+x2+93+c, where ccc is the constant of integration.

Test Index

MHTCET 23 Apr 2025 Shift 1 Paper

Question : 127 of 150

Marks: +1, -0

\int\left(\frac{x-3}{x^{2}+9}\right)^{2} d x=

$\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)-\frac{3}{x^{2}+9}+c$ , where c is the constant of integration.
$\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)-\frac{1}{x^{2}+9}+c$ , where $c$ is the constant of integration.
$\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)+\frac{3}{x^{2}+9}+c$ , where $c$ is the constant of integration.
$\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)-\frac{1}{x^{2}+9}+c$ , where $c$ is the constant of integration.
Let
$\begin{aligned} I & =\int\left(\frac{x-3}{x^{2}+9}\right)^{2} d x \\ & =\int \frac{x^{2}-6 x+9}{\left(x^{2}+9\right)^{2}} d x \\ & =\int\left[\frac{1}{x^{2}+9}-\frac{6 x}{\left(x^{2}+9\right)^{2}}\right] d x \\ & =\int \frac{1}{x^{2}+3^{2}} d x-3 \int \frac{2 x}{\left(x^{2}+9\right)^{2}} d x \end{aligned}$
In $2^{\text {nd }}$ integral,
Put $x^{2}+9=\mathrm{t} \Rightarrow 2 x d x=\mathrm{dt}$
$\begin{aligned} \therefore \quad \mathrm{I} & =\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)-3 \int \frac{\mathrm{dt}}{\mathrm{t}^{2}} \\ & =\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)-3\left(\frac{-1}{\mathrm{t}}\right)+c \\ & =\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)+\frac{3}{x^{2}+9}+c \end{aligned}$

Validate

Solution:

Concept:

Split the integrand into a standard

\tan^{-1}

form and a substitution integral.

Explanation:

First expand:

\int \frac{(x-3)^2}{(x^2+9)^2} dx = \int \frac{x^2 - 6x + 9}{(x^2+9)^2} dx

Separate the terms:

= \int \frac{dx}{x^2+9} - 6\int \frac{x}{(x^2+9)^2} dx

The first integral is

\frac{1}{3}\tan^{-1}\left(\frac{x}{3}\right)

For the second, let

t = x^2 + 9

, so

dt = 2x\,dx

. Then:

-6\int \frac{x}{(x^2+9)^2} dx = -3\int \frac{dt}{t^2} = \frac{3}{t} = \frac{3}{x^2+9}

Thus:

\int\left(\frac{x-3}{x^2+9}\right)^2 dx = \frac{1}{3}\tan^{-1}\left(\frac{x}{3}\right) + \frac{3}{x^2+9} + c

Answer:

Option C:

\frac{1}{3}\tan^{-1}\left(\frac{x}{3}\right)+\frac{3}{x^2+9}+c

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