what is the definite integral of

1)
Upper boundry 4
Lower boundry 0
Problem: x/ square root of 1+2x) dx
2)
Upper boundry 1/2
Lower boundry 0
Problem: sin^-1x / square root of 1-x^2) dx
3)
upper boundry 4
lower boundry 1
Problem: e^square root of x/ square root of x) dx
4)
upper boundry e^4
lower boundry e
Problem: dx/x square root of LN of x)
5) upper boundry pi/2
lower boundry -pi/2
Problem: (x^2 sin x)/ (1+x^6) dx

1)

Upper boundry 4
Lower boundry 0
Problem: x/ square root of 1+2x) dx

x/sqrt(1+2x) = 1/2 2x/sqrt(1+2x) =

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

1/2 [sqrt(1+2x) - 1/sqrt(1+2x)]

Integral of this is:

1/6(1+2x)^(3/2) - 1/2 sqrt(1+2x)

Insert upper and lower boundaries and subtract.

arcsin(x)/sqrt(1-x^2)

1/sqrt(1-x^2) is the derivative of arcsin(x), so the integral is

1/2 arcsin^2(x)

Integral of exp(sqrt(x))/sqrt(x) is

2 exp(sqrt(x))

Integral of 1/[x sqrt(ln(x))] is

2 sqrt[ln(x)]

Problem 5) The function you are integating is odd (i.e. the function changes sign when you change the sign of x), therefore the integral from
-pi/2 to zero will cancel against the integral from zero to pi/2. So, the integral from -pi/2 to pi/2 is zero.

To solve these definite integrals, we can use various integration techniques. Let's go through each problem step by step:

1) Definite integral of (x / sqrt(1+2x)) dx from 0 to 4:

To solve this integral, we can start by simplifying the expression. Multiply the numerator and denominator by sqrt(1+2x) to get x * sqrt(1+2x) / (1+2x).

Now, we can use a technique called u-substitution. Let u = 1+2x. Find du/dx, which is 2. Rearrange the equation to solve for dx, which is du/2.

Now, we can rewrite the integral as: (1/2) * ∫(x * sqrt(u)/u) du.

Integrate the expression ∫(x * sqrt(u)/u) with respect to u. The result is (2/3) * u^(3/2).

Now, plug in the upper and lower boundaries into the antiderivative: [(2/3) * (1+2x)^(3/2)] evaluated from 0 to 4.

Evaluate the expression at x = 4 and subtract the expression at x = 0 to get the final answer.

2) Definite integral of (sin^(-1)x / sqrt(1-x^2)) dx from 0 to 1/2:

We can start by using u-substitution. Let u = sin^(-1)(x), so du = (1 / sqrt(1-x^2)) dx.

Now, we can rewrite the integral as: ∫(u) du.

Integrate the expression ∫(u) with respect to u. The result is (1/2) * u^2.

Now, plug in the upper and lower boundaries into the antiderivative: (1/2) * u^2 evaluated from 0 to sin^(-1)(1/2).

Evaluate the expression at x = sin^(-1)(1/2) and subtract the expression at x = 0 to get the final answer.

3) Definite integral of (e^(sqrt(x)) / sqrt(x)) dx from 1 to 4:

We can start by rewriting the integral as: ∫(e^u / sqrt(u)) du, where u = sqrt(x).

Integrate the expression ∫(e^u / sqrt(u)) with respect to u. The result is 2 * sqrt(u) * e^u.

Now, plug in the upper and lower boundaries into the antiderivative: 2 * sqrt(u) * e^u evaluated from 1 to 4.

Evaluate the expression at x = 4 and subtract the expression at x = 1 to get the final answer.

4) Definite integral of dx / (x * sqrt(ln(x))) from e to e^4:

To solve this integral, we can start by using a substitution. Let u = ln(x). Rearrange the equation to solve for dx, which is du / u.

Now, we can rewrite the integral as: ∫ ((du / u) / (√u)). Simplify the expression: ∫ (du / (u^1.5)).

Integrate the expression ∫ (du / (u^1.5)) with respect to u. The result is -2 / √u.

Now, plug in the upper and lower boundaries into the antiderivative: (-2 / √u) evaluated from e to e^4.

Evaluate the expression at x = e^4 and subtract the expression at x = e to get the final answer.

5) Definite integral of (x^2 * sin(x)) / (1 + x^6) dx from -pi/2 to pi/2:

To solve this integral, we can use the method of partial fractions.

First, factor the denominator (1 + x^6) into (x^2 - x√2 + 1) * (x^2 + x√2 + 1) * (x^2 + 1).

Then, we can use partial fractions to express the integrand as a sum of simpler fractions. For each term in the factorization of the denominator, find the corresponding constant factors.

Once you have the partial fractions representation, integrate each individual term.

Now, plug in the upper and lower boundaries into the antiderivative of each term and subtract the expression at the lower limit from the expression at the upper limit.

This will give you the final answer.