# MAC1105, College Algebra!

## MAC1105, College Algebra!

Write the following expression as a sum and/or difference of logarithms. Express powers as factors.

ln(xsqrt(1+x^(2))) x>0

Thank you very much!!!

ln(xsqrt(1+x^(2))) x>0

Thank you very much!!!

**JFV**- Posts : 10

Join date : 2008-11-18

## Re: MAC1105, College Algebra!

ln(x) + 1/2 * ln(1+(x^2))

**bfrsoccer**- Administrator
- Posts : 61

Join date : 2008-11-09

## Re: MAC1105, College Algebra!

You can solve this question with this method.

The formula for integration by parts is derived by re-arranging the product rule for differentiation.

I[,,f(x)g'(x),x]=f(x)g(x)-I[,,f'(x)g(x…

Let u=f(x) and v=g(x). Then du=f'(x)dx and dv=g'(x)dx, so by the substitution rule, the formula for integration by parts becomes I[,,u,v]=uv-I[,,v,u].

I[,,u,v]=uv-I[,,v,u]

Break the integrand into two separate parts, u and dv.

u=ln(x), dv=(1)/(x^((1)/(2))dx)

Find du by differentiating u, and find v by integrating dv.

u=ln(x), dv=(1)/(x^((1)/(2))dx)_du=(1)/(x) dx, v=2x^((1)/(2))

Replace the values of u, v, and du in the formula for integration by parts.

2x^((1)/(2))ln(x)-I[,,(2)/(x^((1)/(2))…

Solve the second half of the integration by parts formula by finding I[,,(2)/(x^((1)/(2))),x].

I[,,(2)/(x^((1)/(2))),x]

Rewrite the expression to make the exponent of x negative and move it to the numerator.

I[,,2x^(-(1)/(2)),x]

To find the integral of 2x^(-(1)/(2)), find the anti-derivative. The formula for the anti-derivative of a basic monomial is I[x^(n)=(x^(n+1))/((n+1)),x].

4x^((1)/(2))

The indefinite integral also has some unknown constant. This can be proven by completing the opposite operation (derivative) in which C would go to 0. The value of C can be found in cases when an initial condition of the function is given.

4x^((1)/(2))+C

Replace the solved integral back in the formula for integration by parts.

2x^((1)/(2))ln(x)-(4x^((1)/(2)))+C

Simplify the result.

2x^((1)/(2))ln(x)-4x^((1)/(2))+C

Term Papers

The formula for integration by parts is derived by re-arranging the product rule for differentiation.

I[,,f(x)g'(x),x]=f(x)g(x)-I[,,f'(x)g(x…

Let u=f(x) and v=g(x). Then du=f'(x)dx and dv=g'(x)dx, so by the substitution rule, the formula for integration by parts becomes I[,,u,v]=uv-I[,,v,u].

I[,,u,v]=uv-I[,,v,u]

Break the integrand into two separate parts, u and dv.

u=ln(x), dv=(1)/(x^((1)/(2))dx)

Find du by differentiating u, and find v by integrating dv.

u=ln(x), dv=(1)/(x^((1)/(2))dx)_du=(1)/(x) dx, v=2x^((1)/(2))

Replace the values of u, v, and du in the formula for integration by parts.

2x^((1)/(2))ln(x)-I[,,(2)/(x^((1)/(2))…

Solve the second half of the integration by parts formula by finding I[,,(2)/(x^((1)/(2))),x].

I[,,(2)/(x^((1)/(2))),x]

Rewrite the expression to make the exponent of x negative and move it to the numerator.

I[,,2x^(-(1)/(2)),x]

To find the integral of 2x^(-(1)/(2)), find the anti-derivative. The formula for the anti-derivative of a basic monomial is I[x^(n)=(x^(n+1))/((n+1)),x].

4x^((1)/(2))

The indefinite integral also has some unknown constant. This can be proven by completing the opposite operation (derivative) in which C would go to 0. The value of C can be found in cases when an initial condition of the function is given.

4x^((1)/(2))+C

Replace the solved integral back in the formula for integration by parts.

2x^((1)/(2))ln(x)-(4x^((1)/(2)))+C

Simplify the result.

2x^((1)/(2))ln(x)-4x^((1)/(2))+C

Term Papers

**orten999**- Posts : 19

Join date : 2010-04-01

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