To evaluate the integral, we perform the following substitution:

$\displaystyle u=(x+2)^{\frac{1}{2}}$

$\displaystyle du=\frac{1}{2}(x+2)^{-\frac{1}{2}}dx$

The derivative was found using the following rule:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, we rewrite the integrand and integrate:

$\displaystyle 2\int du=2u+C$

The integral was performed using the following rule:

$\displaystyle \int dx=x+C$

Finally, replace u with our original x term:

$\displaystyle 2\sqrt{x+2}+C$

The Laplace Transform will be given by:

$\displaystyle \int_{0}^{\infty}e^{-s(t-a)}*f(t-a) dt$

Since $\displaystyle f(t-a)=0$ when $\displaystyle t \leq a$, we can change the integral to:

$\displaystyle \int_{a}^{\infty}e^{-st}*f(t-a) dt$

This is because when you change the lower bound of the integral, the exponent will only exist for values for which $\displaystyle f$ is defined. 

Let $\displaystyle U=t-a$

$\displaystyle t=U+a$

$\displaystyle \frac{dU}{dt}=1$

This changes our integral to:

$\displaystyle \int_{a}^{\infty}e^{-s(U+a)}*f(U) dU$

We can now move the $\displaystyle e^{-sa}$ term out of the integral, which will give us:

$\displaystyle e^{-sa}\int_{a}^{\infty}e^{-sU}*f(U) dU=e^{-sa} F(s)$

To evaluate this integral, we first make the following substitution:

$\displaystyle u=3x^3+7$

Differentiating this expression, we get:

$\displaystyle du=9x^2 dx\rightarrow x^2dx=\frac{1}{9}du$

Now, we can rewrite the original integral with our substitution and solve:

$\displaystyle \int\frac{18x^2}{(3x^3+7)^2}dx=\int\frac{18}{u^2}\frac{du}{9}=\int2u^{-2}du=-2u^{-1}+C$

Finally, we have to replace u with our earlier definition:

$\displaystyle -2u^{-1}+C=-\frac{2}{3x+7}+C$

Start by substituting $\displaystyle f(b)=e^{kb}$ into the integral to get:

$\displaystyle g(t)=\int_{0}^{t}e^{-s(t-b)}*e^{kb} db$

We can combine this into one large term:

$\displaystyle g(t)=\int_{0}^{t}e^{-s(t-b)+kb} db$

$\displaystyle u=-st+sb+kb$

$\displaystyle \frac{du}{db}=s+k$

$\displaystyle db=\frac{u}{s+k}$

$\displaystyle g(t)=\frac{1}{s+k}\int_{0}^{t}e^{u} du$

Since $\displaystyle \int e^u=e^u$.

$\displaystyle g(t)=\frac{1}{s+k}[e^u]_{0}^{t}=\frac{1}{s+k}[e^{-st+sb+kb}]_{0}^{t}= \frac{1}{s+k}[e^{kt}-e^{-st}]$

This can only grow when: 

$\displaystyle k > 0$ 

$\displaystyle \small \small \int\frac{ \sin(\ln x)}{x} dx$

Let

$\displaystyle \small u = ln(x)$

$\displaystyle \small du =\frac{1}{x}dx$

$\displaystyle \small \small \small \small \int\frac{ \sin(\ln x)}{x} dx=\int\sin(u)du=-\cos(u)$

We can now convert this back to a function of $\displaystyle \small x$ by substituting $\displaystyle \small u = \ln x$, 

$\displaystyle \small -\cos(u)=-\cos(\ln x)$

$\displaystyle \small \small \small \small \small \int\frac{ \sin(\ln x)}{x} dx = - \cos(\ln x)$

$\displaystyle \int \frac{x}{x+2}dx$

Add 2 and subtract 2 from the numerator of the integrand:$\displaystyle \int \frac{x}{x+2}dx=\int \frac{x+2-2}{x+2}dx$.

Simplify and apply the difference rule:$\displaystyle \int \frac{x+2-2}{x+2}dx=\int (\frac{x+2}{x+2}-\frac{x}{x+2})dx=\int dx-\int \frac{2}{x+2}dx$

Solve the first integral: $\displaystyle \int dx=x+C$.

Make the following substitution to solve the second integral: $\displaystyle u=x+2$ $\displaystyle du=dx$

Apply the substitution to the integral: $\displaystyle \int \frac{2}{x+2}dx=2\int\frac{du}{u}$

Solve the integral:$\displaystyle 2\int\frac{du}{u}=2ln\left | x+2\right |+C$

Combine the answers to the two integrals: $\displaystyle \int dx-\int \frac{2}{x+2}dx=x-2ln\left | x+2\right |+C$.

Solution: $\displaystyle \int \frac{x}{x+2}dx=x-2ln\left | x+2\right |+C$

We use substitution to solve the problem:

Let  $\displaystyle u=2+cos(x)$  and  $\displaystyle du=-sin(x)dx$  $\displaystyle \Rightarrow sinxdx=-du$

Therefore:

$\displaystyle \int \frac{sin(x)}{2+cos(x)}dx=-\int \frac{1}{u}du=-ln(u)+c=-ln(2+cos(x))+c$

Here we use substitution to solve for the integrand.  Let u=sin(x) therefore du= cos(x)dx.  Plug your values back in:

$\displaystyle \int cos(x)e^{sinx}dx = \int e^udu = e^u +c =e^{sinx}+c$

To integrate this expression, you have to use u substitution. First, assign your u expression:

$\displaystyle u=x^2+1$

$\displaystyle du=2xdx$

$\displaystyle \frac{1}{2}du=xdx$

Now, plug everything back in so you can integrate:

$\displaystyle \frac{1}{2}\int_{1}^{2}\frac{1}{u}du$

Now integrate:

$\displaystyle \frac{1}{2}\ln\left | u \right |$

From here substitute the original variable back into the expression.

$\displaystyle \frac{1}{2}\ln|x^2+1|$

Evaluate at 2 and then 1.

Subtract the results:

$\displaystyle \frac{1}{2}\left(\ln|2^2+1|-\ln|1+1| \right )$

$\displaystyle \frac{1}{2}\left(\ln|5|-\ln|2|\right)$

$\displaystyle \frac{\ln5-\ln2}{2}=0.4581$

$\displaystyle y=\int \sqrt{4-4x^{2}}dx$

Factor out $\displaystyle \sqrt{4}$ from the integrand, and simplify: 

$\displaystyle y=\int \sqrt{4}\sqrt{1-x^{2}}dx=\int 2\sqrt{1-x^{2}}dx=2\int \sqrt{1-x^{2}}dx$

Make the following substitution: $\displaystyle x=sin\theta$ $\displaystyle dx=cos\theta d\theta$

Plug the substitution into the integrand: 

$\displaystyle y=2\int \sqrt{1-sin^{2}\theta }cos\theta d\theta$.

Use the Pythagorean identity to make the following substitution, and simplify: 

$\displaystyle \\y=2\int \sqrt{1-sin^{2}\theta }cos\theta d\theta\\=2\int \sqrt{cos^{2}\theta }cos\theta d\theta\\=2\int cos\theta cos\theta d\theta\\= 2\int cos^{2}\theta d\theta$

Apply the following identity to the integrand: 

$\displaystyle cos^{2}\theta=\frac{1+cos(2 \theta )}{2}$: 

$\displaystyle y=2\int cos^{2}\theta=2\int \frac{1+cos(2) \theta }{2}d\theta$.

Separate the integral into two separate integrals: 

$\displaystyle y= 2\int \frac{1}{2}d\theta+2\int\frac{cos(2\theta )}{2}d\theta$.

Solve the first integral: 

$\displaystyle y=\theta+2\int\frac{cos(2\theta )}{2}d\theta$.

Make the following substitution for the second integral: 

$\displaystyle u=2\theta$ $\displaystyle du=2d\theta$ $\displaystyle \frac{1}{2}du=d\theta$.

Apply the substitution, and solve the integral: 

$\displaystyle 2\int \frac{cos(2\theta )}{2}=2\int \frac{cosu}{2}\frac{1}{2}du=\frac{1}{2}sinu=\frac{1}{2}sin(2\theta )+C$.

Combine answers for both integrals: 

$\displaystyle y=2\int cos^{2}\theta=\theta+\frac{1}{2}sin(2\theta )+C$

Solve for $\displaystyle \theta$: 

$\displaystyle x=sin\theta$ 

$\displaystyle sin^{-1}x=\theta$

Plug values for  $\displaystyle \theta$ back into solution to integral: 

Recall that, 

$\displaystyle sin(2\theta)=2sin(\theta)cos(\theta)$

and from above,

$\displaystyle \\x=sin(\theta) \\cos(\theta)=\sqrt{1-sin^2(\theta)}=\sqrt{1-x^2}$

Therefore,

$\displaystyle \\\theta+\frac{1}{2}sin(2\theta )+C \\=sin^{-1}+4sin(x)cos(x) \\=sin^{-1}(x)+x\sqrt{1-x^2}+C$.