Question

How do you find the intergal of (x^2+3x)^2/(2x+3)?

Answer 1

The answer is `=1/128(2x+3)^4-9/32(2x+3)^2+81/32ln(|2x+3|)+C`

Explanation:

Perform this integral by substitution

Let `u=2x+3`, `=>`, `du=2dx`

`x=(u-3)/2`

`x^2=(u-3)^2/4`

`x^2+3x=(u-3)^2/4+3*(u-3)/2=(u-3)/4(u-3+6)=(u-3)(u+3)/4`

Therefore,

`int((x^2+3x)^2dx)/(2x+3)=1/32int((u-3)^2(u+3)^2du)/u`

`=1/32int((u^2-9)^2du)/u`

`=1/32int((u^4-18u^2+81)du)/u`

`=1/32int(u^3-18u+81/u)du`

`=1/128*u^4-9/32u^2+81/32lnu`

`=1/128(2x+3)^4-9/32(2x+3)^2+81/32ln(|2x+3|)+C`

Answer 2

`1/128[(2x+3)^4-36(2x+3)^2+324ln|(2x+3)|]+C, or,`

`1/128{(2x+3)^2(2x+9)(2x-3)+324ln|(2x+3)|}+C`.

Explanation:

Suppose that, `I=int(x^2+3x)^2/(2x+3)dx`.

Subst. `2x+3=y rArr x=(y-3)/2, and, :., dx=1/2dy`.

Further, `(x^2+3x)^2/(2x+3)=1/(2x+3)*{x(x+3)}^2`,

`=1/y*{((y-3)/2)((y-3)/2+3)}^2`,

`=1/y{((y-3)/2)((y+3)/2)}^2`,

`=1/y*{(y^2-9)/4}^2`,

`=1/y((y^4-18y^2+81)/16)`,

`=1/16(y^4/y-18y^2/y+81/y)`.

`:. I=1/16int(y^3-18y+81/y)*1/2dy`,

`=1/32(y^4/4-18*y^2/2+81ln|y|)`,

`=1/32(y^4/4-9y^2+81ln|y|)`.

`rArr I=1/128[(2x+3)^4-36(2x+3)^2+324ln|(2x+3)|]+C`

or,

`I=1/128{(2x+3)^2(2x+9)(2x-3)+324ln|(2x+3)|}+C`.

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Answer 3

`1/8x^4+3/4x^3+9/16x^2-27/16x+81/32ln|(2x+3)|+C`.

Explanation:

Have a look at this Second Method :

By Long Division, we get,

`(x^2+3x)^2/(2x+3)`,

`=1/2x^3+9/4x^2+9/8x-27/16+81/(16(2x+3))`.

`:. int(x^2+3x)^2/(2x+3)dx`,

`=1/2*x^4/4+9/4*x^3/3+9/8*x^2/2-27/16*x`

`+81/16*1/2ln|(2x+3)|`,

`=1/8x^4+3/4x^3+9/16x^2-27/16x+81/32ln|(2x+3)|+C`.

Answer 4

`I=x^4/8+(3x^3)/4+(9x^2)/16-(27x)/16+81/32ln|2x+3|+c`

Explanation:

Here,

`I=int(x^2+3x)^2/(2x+3)dx`

`=int(x^4+6x^3+9x^2)/(2x+3)dx`

We have ,

Dividend `= x^4+6x^3+9x^2 and` Divisor `=2x+3`

#"Using the method of " color(blue)"Long Division of Polynomial"#

we get , Quotient `=(x^3)/2+(9x^2)/4+(9x)/8-27/16`

`and` Remainder `=81/16`

So,

`I=int((x^3)/2+(9x^2)/4+(9x)/8-27/16+(81/16)/(2x+3))dx`

`=x^4/8+9/4*x^3/3+(9x^2)/16-(27x)/16+81/16*1/2ln|2x+3|+c`

`I=x^4/8+(3x^3)/4+(9x^2)/16-(27x)/16+81/32ln|2x+3|+c`