`int_0^1 (x^3 + 2x)/(x^4 + 4x^2 + 3) dx` Evaluate the integral

`int_0^1 (x^3+2x)/(x^4+4x^2+3)dx`

To solve, apply u-substitution method.

Let

`u = x^4+4x^2+3`

Then, differentiate u.

`du=(4x^3 + 8x)dx`

`du=4(x^3+2x)dx`

`(du)/4=(x^3+2x)dx`

And, determine the values of u when x=0 and x=1.

`x=0`

`u=0^4+4(0)^2+3=3`

`x=1`

`u=1^4+4(1)^2+3=8`

So expressing the integral in terms of u it becomes:

`int_0^1(x^3+2x)/(x^4+4x^2+3)dx `

`= int_3^8 1/u * (du)/4 `

`= 1/4int_3^8 1/u du`

`= 1/4 ln u |_3^8`

`= 1/4 (ln 8 - ln 3)`

`=0.2452`

Therefore, `int_0^1 (x^3+2x)/(x^4+4x^2+3)dx = 0.2452` .

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