# What is integral(0 down - 1 up) f(square root x) dx? f(x) = ln(1+x ^2)

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### 1 Answer

You need to evaluate `f(sqrt x)` , hence, you need to replace `sqrt x` for x in equation of functino, such that:

`f(sqrt x) = ln(1 + sqrt(x^2))`

Using the absolute value definition, yields:

`sqrt x^2 = |x| = {(x, x>=0),(-x, x< 0):}`

Since the limits of integration are 0 and 1, hence, you need to replace `x` for `sqrt x^2` in equation of function.

`f(sqrt x) = ln(1 + x)`

Hence, you need to evaluate the definite integral, such that:

`int_0^1 f(sqrt x)dx =int_0^1 ln(1 + x) dx`

You need to use integration by parts, such that:

`u = ln(1 + x) => du = 1/(1+x)dx`

`dv = dx => v = x`

`int_a^b udv = uv|_a^b - int_a^b vdu`

Reasoning by analogy, yields:

`int_0^1 ln(1 + x) dx = x*ln(1 + x)|_0^1 - int_0^1 x/(1 + x)dx`

You need to evaluate the definite integral `int_0^1 x/(1 + x)dx` such that:

`int_0^1 x/(1 + x)dx = int_0^1 (x + 1 - 1)/(1 + x)dx`

Using the property of linearity of integral yields:

`int_0^1 (x + 1 - 1)/(1 + x)dx = int_0^1 (x + 1)/(1 + x)dx - int_0^1 1/(1 + x)dx`

`int_0^1 (x + 1 - 1)/(1 + x)dx = int_0^1 dx - int_0^1 1/(1 + x)dx`

`int_0^1 (x + 1 - 1)/(1 + x)dx = x|_0^1 - ln(x+1)|_0^1`

`int_0^1 ln(1 + x) dx = x*ln(1 + x)|_0^1 - x|_0^1 + ln(x+1)|_0^1`

Using the fundamental theorem of calculus, yields:

`int_0^1 ln(1 + x) dx = 1*ln2 - 0ln 1 - 1 + 0 + ln 2 - ln 1`

`int_0^1 ln(1 + x) dx = ln 2 - 1 + ln 2`

`int_0^1 ln(1 + x) dx = 2ln 2 - 1`

`int_0^1 ln(1 + x) dx = ln 2^2 - ln e`

Converting the difference of logarithms into the logarithm of quotient, yields:

`int_0^1 ln(1 + x) dx = ln(4/e)`

**Hence, evaluating the given definite integral, using integration by parts, yields **`int_0^1 ln(1 + x) dx = ln(4/e).`