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Integrate (2+3x)/(3-2x) kindly show the answer …

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Integrate (2+3x)/(3-2x) kindly show the answer step by step.

Posted by Srinivas Chandra 6 years, 10 months ago

CBSE > Class 12 > Mathematics

1 answers

Naveen Sharma 6 years, 10 months ago

Ans. {tex}\int{ 2+3x\over 3-2x}dx{/tex}

As this is not a proper fraction. then first we have to conert it in proper fraction.

{tex}=> \ \int ({ -{3\over 2} + {13\over 2}{1\over (-2x+3)} }) dx {/tex}

{tex}=> \ \int { -{3\over 2} }dx + \int {{13\over 2}{1\over (-2x+3)}} dx {/tex}

{tex}=> \ -{3\over 2} \int dx + {13\over 2 }\int {1\over -2x+3} dx {/tex}
Multiply and divide integral by -2, we get

{tex}=> \ -{3\over 2} \int dx - {13\over 4 }\int {-2\over -2x+3} dx {/tex}

{tex}=> \ -{3\over 2} x - {13\over 4 }I_2 \ \ \ ........... (1){/tex}

{tex}I_2 = \int {-2\over -2x+3} dx {/tex}

Put -2x + 3 = t then

-2 dx = dt

So,

{tex}I_2 = \int {1\over t}dt{/tex}

=> I₂ = log t = log (-2x+3)

Put in (1). we get

{tex}=> \ -{3\over 2} x - {13\over 4 }log (-2x+3){/tex}

1Thank You

ANSWER

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