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Integrate the following functions w.r.t. x : 7+4+5x2(2x+3)32

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प्रश्न

Integrate the following functions w.r.t. x : `(7 + 4 + 5x^2)/(2x + 3)^(3/2)`

योग
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उत्तर

Let I = `int(7 + 4x +5x^2)/(2x + 3)^(3/2).dx`

= `int(5x^2 + 4x + 7)/(2x + 3)^(3/2).dx`

Put 2x + 3 = t

∴ 2dx = dt

∴ dx = `dt/(2)`

Also, x = `(t - 3)/(2)`

∴ I = `int(5((t - 3)/2)^2 + 4((t - 3)/2) + 7)/t^(3/2).dt/(2)`

= `(1)/(2) int(5((t^2 -  6t  +  9)/4) + 2(t - 3) + 7)/t^(3/2)dt`

= `(1)/(2)int (5t^2 - 30t + 45 + 8t - 24 + 28)/(4t^(3/2))dt`

= `(1)/(8)int(5t^2 - 22t + 49)/t^(3/2)dt`

= `(1)/(8)int(5t^(1/2) - 22t^(-1/2) + 49t^(-3/2))dt`

= `(5)/(8)intt^(1/2)dt - 22/8 int t^(-1/2)dt + 49/8 int t^(-3/2)dt`

= `(5)/(8).t^(3/2)/((3/2)) - (11)/(4).t^(1/2)/((1/2))  + (49)/(8).t^(-1/2)/((-1/2)) + c`

= `(5)/(12)(2x+ 3)^(3/2) - (11)/(2)sqrt(2x + 3) - (49)/(4).(1)/sqrt(2x + 3) + c`.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Indefinite Integration - Exercise 3.2 (A) [पृष्ठ ११०]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Indefinite Integration
Exercise 3.2 (A) | Q 1.22 | पृष्ठ ११०

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