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

Expert Answers
Neethu eNotes educator| Certified Educator

We have to evaluate the integral `\int \frac{xdx}{(2x^2+3)^5}`

Let us use the substitution:


Differentiating the above, we get,



Therefore, we can write the integral as,

`\int \frac{xdx}{2x^2+3}=\frac{1}{4}\int \frac{dt}{t^5}`

              `=\frac{1}{4}\int t^{-5}dt` ``

               `=\frac{1}{4}.\frac{t^{-5+1}}{-5+1}+C` where C is a constant.


                 `=\mathbf{\frac{-1}{16(2x^2+3)^4}}` +C

mathewww | Student

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:




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


steamgirl | Student

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)

simplelucker | Student

To find 

Let  , then  . The substitution gives:

Applying the formula  we get:

Replacing  with  , we get the required solution as:

kailash | Student


We can asnwer this integration by method of substitution.







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




where c is integrating constant.

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