∫ {x}{{x+1}+{x-1}} dx = A (x) (x+1)³/2} + B (x) (x-1)³/2} + C , then A (x) + B (x) =

Correct Answer is : - 4154/15154

Correct Answer is : - 4154/15154

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AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 72 of 160

Marks: +1, -0

∫

\frac{x}{\sqrt{x+1}+\sqrt{x-1}}

(x+1)^{3/2}

(x-1)^{3/2}

+ C , then A (x) + B (x) =

Solution:

Consider the expression,

I=\int \frac{x}{\sqrt{x+1}+\sqrt{x-1}} \, dx

=\int \frac{x[\sqrt{x+1}-\sqrt{x-1}]}{x+1-x+1} \, dx

= \frac{1}{2} \int x \sqrt{x+1} \, dx - \frac{1}{2} \int x \sqrt{x-1} \, dx

= \frac{1}{2} I_{1} - \frac{1}{2} I_{2}

Then,

I_{1}= \int x \sqrt{x+1} \, dx

Put

x+1=u \Rightarrow dx=du

I_{1}= \int (u-1) \sqrt{u} \, dx

= \int (u^{3/2} - u^{1/2}) \, dx

= \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} + c_{1}

= \frac{2}{5} (x+1)^{5/2} - \frac{2}{3} (x+1)^{3/2} + c_{1}

Further simplify the above,

I_{1}=2(x+1)^{3/2} \left[ \frac{3x+3-5}{15} \right] + c_{1}

= \frac{2(3x-2)}{15} (x+1)^{3/2} + c_{1}

Now,

I_{2}= \int x \sqrt{x-1} \, dx

Put

x-1=v \Rightarrow dx=dv

I_{2}= \int (v+1) \sqrt{v} \, dv

= \int (v^{3/2} + v^{1/2}) \, dv

= \frac{2}{5} v^{5/2} + \frac{2}{3} v^{3/2} + c_{2}

=2(x-1)^{3/2} \left[ \frac{1}{5}(x-1) + \frac{1}{3} \right] + c_{2}

=2(x-1)^{3/2} \left[ \frac{3x-3+5}{15} \right] + c_{2}

Further simplify the above,

I_{2}= \frac{2(3x+2)}{15} (x-1)^{3/2} + c_{2}

This gives,

I= \frac{1}{2} I_{1} - \frac{1}{2} I_{2}

= \frac{(3x-2)}{15} (x+1)^{3/2} - \frac{(3x+2)}{15} (x-1)^{3/2} + c

[\because c=c_{1}+c_{2}]

This implies,

A(x)= \frac{1}{15}(3x-2)

and

B(x)= \frac{-1}{15}(3x+2)

So,

A(x)+B(x)= \frac{1}{15}(3x-2) - \frac{1}{15}(3x+2)

= -\frac{4}{15}

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