\(\displaystyle \int{6x\ln\left(\frac{7}{2}x^2+3\right)}dx\)                                                       (1)

1) Simplify with a substitution. 

It is often necessary to define a new variable \(\displaystyle u\), carefully chosen so that rewriting the integrand in terms of this new variable will make integration easier. In this case, the obvious variable to introduce will be defined by, 

\(\displaystyle u =\frac{7}{2}x^2+3\)                                                                      (2)

\(\displaystyle du=7xdx\)                                                                         (3)

Use equations (2) and (3) to rewrite (1).  

\(\displaystyle \int{6x\ln\left(\frac{7}{2}x^2+3\right)}dx=\frac{6}{7}\int{\ln\left(\frac{7}{2}x^2+3\right)} \underset{du}{ \underbrace{7xdx}}=\frac{6}{7}\int\ln udu\)

 2) Use integration by parts

To compute \(\displaystyle \int \ln udu\) use integration by parts. Ignore the constant out front for the moment, 

\(\displaystyle \int vw'=vw-\int v'w\)                                                 (4)

Define \(\displaystyle v\) and \(\displaystyle w\) and insert into the equation (4). 

 \(\displaystyle v=\ln u\) 

 \(\displaystyle w = u\),

 \(\displaystyle v'=\frac{1}{u}du\)

\(\displaystyle w'=du\)

 \(\displaystyle \int\ln udu=u\ln u-\int u\left(\frac{1}{u}du \right )\)

\(\displaystyle \int\ln udu=u\ln u-u+C\)                                               (5)

Let's factor the non-constant terms in equation (5), this will make the result easier to express when we convert back to \(\displaystyle x.\)

\(\displaystyle [u(\ln u-1)]+C\)

We previously had a constant in front of the integral, 

\(\displaystyle \frac{6}{7}\int\ln(u)du =\frac{6}{7}[u(\ln u-1)]+C\) 

Now we can write the integral terms of the original variable \(\displaystyle x\) by substituting equation (2) into the previous expression to obtain, 

\(\displaystyle =\frac{6}{7}\left(\frac{7}{2}x^2+3\right )\left[\ln\left(\frac{7}{2}x^2+3\right)-1\right]+C\)

\(\displaystyle =\left(3x^2+\frac{18}{7}\right)\left[\ln\left(\frac{7}{2}x^2+3 \right )-1\right] +C\)

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:

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

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

\(\displaystyle \frac{\ln5-\ln2}{2}=0.4581\)