Correct Answer is : 121/221 log (∣x2+2x∣(x+1)2)≤ft(|x²+2x|/(x+1)²)((x+1)2∣x2+2x∣) + c

log |x| + log ≤ft|x+2/x+1| + c

log |x| - log |x + 1| + log |x + 2| + c

1/2 [log |x| - log |x + 1| + log |x + 2|] + c

1/2 log ≤ft(|x²+2x|/(x+1)²) + c

Correct Answer is : 121/221 log (∣x2+2x∣(x+1)2)≤ft(|x²+2x|/(x+1)²)((x+1)2∣x2+2x∣) + c

Test Index

AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

Section: Mathematics

Question : 74 of 160

Marks: +1, -0

∫

\frac{dx}{x^3+3x^2+2x}

Solution:

It is assumed.

I=\int\frac{dx}{x^1+3x^2+2x}

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

=\int\frac{dx}{x(x+1)(x+2)}

Let

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

+\frac{C}{(x+2)}

1=A(x+1)(x+2)+Bx(x+2)

+Cx(x+1)

Substitute the value of

x=-1

the value of

B

will be -1

B=-1

Substitute

x=-2

we get

C=\frac{1}{2}

I=\int\left(\frac{1}{2x}-\frac{1}{x+1}+\frac{1}{2(x+2)}\right)dx

After differentiate the above expression

=\frac{1}{2}\log x-\log(x+1)

+\frac{1}{2}\log(x+2)+C

=\frac{1}{2}[\log x-2\log(x+1)

+\log(x+2)]+C

=\frac{1}{2}\log\left|\frac{x(x+2)}{(x+1)^2}\right|+C

=\frac{1}{2}\log\left|\frac{x^2+2x}{(x+1)^2}\right|+C

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