Evaluate: x/(x+2)(3-2x) dx

Correct Answer is : −27log⁡∣x+2∣−314log⁡∣3−2x∣+C-2/7 |x+2| - 3/14 |3-2x| + C−72log∣x+2∣−143log∣3−2x∣+C

Correct Answer is : −27log⁡∣x+2∣−314log⁡∣3−2x∣+C-2/7 |x+2| - 3/14 |3-2x| + C−72log∣x+2∣−143log∣3−2x∣+C

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UPSC NDA Math Model Paper 3

Question : 50 of 120

Marks: +1, -0

Evaluate:

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

Solution:

Here we have to find the value of

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

\Rightarrow \frac{x}{(x+2)(3-2x)} = \frac{A}{x+2}

+ \frac{B}{3-2x}

\Rightarrow x = A (3 - 2x) + B (x + 2)

........(1)

By putting x = - 2 on both the sides of (1) we get A = - 2/7

By putting x = 3/2 on both the sides of (1) we get B = 3/7

\Rightarrow \int \frac{x}{(x+2)(3-2x)} dx = -\frac{2}{7} \int \frac{dx}{x+2}

+ \frac{3}{7} \int \frac{dx}{3-2x}

As we know that

\int \frac{dx}{x} = \log |x| + C

where

C

is a constant

\Rightarrow \int \frac{x}{(x+2)(3-2x)} dx

= -\frac{2}{7} \log |x+2| - \frac{3}{14} \log |3-2x| + C

where

C

is a constant

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