If x ≠ {-3}{{2}} , then ∫ {x²{2x²+6{2}+9} dx =

Correct Answer is : 222[(2x+3)−6log⁡∣2x+3∣−92x+3]{2}{2{2}} ≤ft[ ({2}x+3) - 6 ≤ft| {2}x+3 | - {9}{{2}x+3} ]222[(2x+3)−6log2x+3−2x+39] + c

{2}{2{2}} ≤ft[ ({2}x+3) - 6 ≤ft| {2}x+3 | - {9}{{2}x+3} ] + c

{2}{2{2}} ≤ft[ {2}x+3 - 6 ≤ft| ({2}x+3) | + {9}{{2}x+3} ] + c

Correct Answer is : 222[(2x+3)−6log⁡∣2x+3∣−92x+3]{2}{2{2}} ≤ft[ ({2}x+3) - 6 ≤ft| {2}x+3 | - {9}{{2}x+3} ]222[(2x+3)−6log2x+3−2x+39] + c

Test Index

AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

Section: Mathematics

Question : 72 of 160

Marks: +1, -0

If x ≠

\frac{-3}{\sqrt{2}}

\frac{x^2}{2x^2+6\sqrt{2}+9}

dx =

$\frac{2}{2\sqrt{2}} \left[ (\sqrt{2}x+3) - 6 \log \left| \sqrt{2}x+3 \right| - \frac{9}{\sqrt{2}x+3} \right]$
+ c
$\frac{2}{2\sqrt{2}} \left[ \sqrt{2}x+3 - 6 \log \left| (\sqrt{2}x+3) \right| + \frac{9}{\sqrt{2}x+3} \right]$
+ c
$\sqrt{2}$ x + 3 - 6 log $(\sqrt{2}x+3)$ + c
log $(2x^2+6\sqrt{2}x+9)$ + c

Solution:

I = \int \frac{x^2}{2x^2+6\sqrt{2}x+9} \, dx

It is solved as

I \; = \int \frac{\frac{1}{2}(2x^2+6\sqrt{2}x+9) - \left(3\sqrt{2}x+\frac{9}{2}\right)}{2x^2+6\sqrt{2}x+9} \, dx

\; = \int \frac{1}{2} \, dx - \int \frac{3\sqrt{2}x+\frac{9}{2}}{2x^2+6\sqrt{2}x+9} \, dx

\; = \frac{x}{2} - \int \frac{3\sqrt{2}x+\frac{9}{2}}{2x^2+6\sqrt{2}x+9} \, dx

Now

3\sqrt{2}x + \frac{9}{2} = A(4x+6\sqrt{2}) + B

Compare the coefficients of like terms.

4A = 3\sqrt{2}

A = \frac{3\sqrt{2}}{4}

And

6\sqrt{2}A + B = \frac{9}{2}

B = -\frac{9}{2}

Therefore

3\sqrt{2}x + \frac{9}{2} = 3 \frac{\sqrt{2}}{4} (4x+6\sqrt{2}) + \frac{9}{4}

Hence.

I = \frac{x}{2} - \int \frac{3\frac{\sqrt{2}}{4}(4x+6\sqrt{2}) + \frac{9}{4}}{2x^2+6\sqrt{2}x+9} \, dx

\; = \frac{x}{2} - 3\frac{\sqrt{2}}{4} \int \frac{4x+6\sqrt{2}}{2x^2+6\sqrt{2}x+9} \, dx + \frac{9}{4} \int \frac{1}{2x^2+6\sqrt{2}x+9} \, dx

\; = \frac{x}{2} - \frac{3\sqrt{2}}{4} \log \left| 2x^2+6\sqrt{2}x+9 \right| + \frac{9}{2} \int \frac{1}{(\sqrt{2x}+3)^2} \, dx

\; = \frac{x}{2} - \frac{3\sqrt{2}}{4} \log \left| (\sqrt{2x}+3)^2 \right| + \frac{9}{2} \frac{(\sqrt{2x}+3)^{-1}}{-1} \cdot \frac{1}{\sqrt{2}} + c'

Solve further

I = \frac{x}{2} - \frac{3\sqrt{2}}{2} \log \left| \sqrt{2x}+3 \right| + \frac{9}{2\sqrt{2}} \frac{(\sqrt{2x}+3)^{-1}}{-1} + c'

\; = \frac{1}{2\sqrt{2}} \left[ (\sqrt{2}x+3) - 6 \log \left| \sqrt{2}x+3 \right| - \frac{9}{\sqrt{2}x+3} \right] + c

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