`int(x dx )/ (2x^2 +3)^5` Evaluate the integral

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mathewww | High School Teacher | (Level 1) Honors

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To find `int (xdx)/(2x^2+3)^5`

Let `u=2x^2+3` , then `du=4x dx rArr 1/4du=xdx` . The substitution gives:

`int (xdx)/(2x^2+3)^5`

`=int 1/(4u^5) du`

`=int 1/4*u^-5 du`

Applying the formula `int x^n dx = x^(n+1)/(n+1)+C` we get:

`1/4*u^(-5+1)/(-5+1)+C`

`=1/4*u^-4/-4+C`

`=-1/(16u^4)+C`

Replacing `u` with `2x^2+3` , we get the required solution as:

`-1/(16(2x^2+3)^4)+C.`

Sources:
steamgirl's profile pic

steamgirl | Student, College Junior | (Level 1) Honors

Posted on

For this problem you want to use U-substitution. Set u = to 2x^2 + 3, therefore du = 4x dx, or du/4 = x dx.

(1/4) the integral of u^-5 du.

Which with the power rule, would mean you add one to the exponent and divide by the resulting number. This gives you:

(-1/16)(1/(2x^2 + 3)^4)

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simplelucker | (Level 1) eNoter

Posted on

To find 

Let  , then  . The substitution gives:

Applying the formula  we get:

Replacing  with  , we get the required solution as:

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kailash | Elementary School Teacher | eNotes Newbie

Posted on

`int(xdx)/(2x^2+3)^5`

We can asnwer this integration by method of substitution.

`t=2x^2+3`

`dt=4xdx`

`(1/4)dt=xdx`

Thus,

`int(xdx)/(2x^2+3)^5=int(dt/4)(1/t^5)`

`=(1/4)intdt/t^5`

`=(1/4)int t^(-5)dt`

`=(1/4)t^(-5+1)/(-5+1)+c`

`=(-1/16)t^(-4)+c`

`=-1/16(2x^2+3)^(-4)+c`

where c is integrating constant.

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