# Please explain how would calculate integral `int_0^sqrt7` x^3/(1+x^2)^(1/2)dx?

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### 1 Answer

You need to evaluate the given definite integral, hence, you need to perform the following steps on integrand, such that:

(x + x^3 - x)/(sqrt(1 + x^2)) = (x + x^3)/(sqrt(1 + x^2)) - x/(sqrt(1 + x^2))

(x + x^3 - x)/(sqrt(1 + x^2)) = x(1 + x^2)/(sqrt(1 + x^2)) - x/(sqrt(1 + x^2))

Reducing duplicate factors yields:

(x + x^3 - x)/(sqrt(1 + x^2)) = xsqrt(1 + x^2) - x/(sqrt(1 + x^2))

Integrating both sides, yields:

int_0^sqrt7 (x^3)/(sqrt(1 + x^2)) dx = int_0^sqrt7 x sqrt(1 + x^2) dx - int_0^sqrt7 x/(sqrt(1 + x^2)) dx

You need to evaluate the definite integral `int_0^sqrt7 x sqrt(1 + x^2) dx` using substitution, such that:

`x^2 + 1 = t => 2xdx = dt`

`int x sqrt(1 + x^2) dx = int sqrt t*(dt)/2 => int sqrt t*(dt)/2 = (1/2) t^(1/2+1)/(1/2+1)`

`int_0^sqrt7 x sqrt(1 + x^2) dx = (1/3)(x^2 + 1)sqrt(x^2 + 1)|_0^sqrt7`

Using the fundamental theorem of calculus, yields:

`int_0^sqrt7 x sqrt(1 + x^2) dx = (1/3)(7 + 1)sqrt(7 + 1) - 1/3`

`int_0^sqrt7 x sqrt(1 + x^2) dx = (1/3)(8sqrt8 - 1)`

`int_0^sqrt7 x sqrt(1 + x^2) dx = (16sqrt2 - 1)/3`

You need to evaluate the definite integral `int_0^sqrt7 x/(sqrt(1 + x^2)) dx` using the same substitution, such that:

`x^2 + 1 = t => 2xdx = dt`

`int x/(sqrt(1 + x^2)) dx = int t^(-1/2) (dt)/2`

`int x/(sqrt(1 + x^2)) dx = (1/2) t^(-1/2+1)/(-1/2)`

`int_0^sqrt7 x/(sqrt(1 + x^2)) dx = -sqrt (1 + x^2)|_0^sqrt7`

`int_0^sqrt7 x/(sqrt(1 + x^2)) dx = -sqrt 8 + sqrt1`

`int_0^sqrt7 x/(sqrt(1 + x^2)) dx= 1 - 2sqrt2`

`int_0^sqrt7 (x^3)/(sqrt(1 + x^2)) dx = (16sqrt2 - 1)/3 + 2sqrt2 - 1`

`int_0^sqrt7 (x^3)/(sqrt(1 + x^2)) dx = (16sqrt2 + 6sqrt2 - 1 - 3)/3`

`int_0^sqrt7 (x^3)/(sqrt(1 + x^2)) dx = (22sqrt2 - 4)/3`

**Hence, evaluating the given definite integral, under the given conditions, yields **`int_0^sqrt7 (x^3)/(sqrt(1 + x^2)) dx = (22sqrt2 - 4)/3.`