To solve this problem we need to use u-substitution. The key to knowing that is by noticing that we have both an \(\displaystyle x\) and an \(\displaystyle x^2\) term, and that hypothetically if we could take the derivate of the \(\displaystyle x^2\) term it could cancel out the \(\displaystyle x\) term. Let's take a closer look.

Let's choose our \(\displaystyle u\) in this problem to be \(\displaystyle x^2\). We would need to calculate the \(\displaystyle du\) term since we're switching from the x variable to the u variable.

\(\displaystyle u=x^2\)

\(\displaystyle \frac{du}{dx}=2x,dx=\frac{du}{2x}\)

Let's take a look back at the orginal problem and begin to make substitutions.

\(\displaystyle \int xsin(x^2)dx=\int xsin(u)\frac{du}{2x}\)

Notice here that the x's will cancel out, leaving us with an integral with entirely the u variable.

\(\displaystyle \frac12 \int sin(u)du\)

This is a nice simple integral that results in

\(\displaystyle -\frac12 cos(u)+c\). Now all we need to do is replace that u with the original variable.\(\displaystyle u=x^2\)

\(\displaystyle -\frac12 cos(x^2)+c\)

The problem \(\displaystyle \int \frac{2x-3}{2x^2-6x}dx\) is a U-substitution question. The term \(\displaystyle u\) might not be easily seen, but the \(\displaystyle du\) term must equal to \(\displaystyle 2x-3\: dx\).

Factor the denominator by taking \(\displaystyle 2\) as the common factor.

\(\displaystyle 2x^2-6x= 2(x^2-3x)\)

Rewrite the integral.

\(\displaystyle \int \frac{2x-3}{2x^2-6x}dx =\int \frac{2x-3}{2(x^2-3x)}dx=\frac{1}{2}\int \frac{2x-3}{(x^2-3x)}dx\)

\(\displaystyle u=x^2-3x\)

\(\displaystyle du=2x-3\: dx\)

\(\displaystyle \frac{1}{2}\int\frac{1}{u}du=\frac{1}{2}ln\left | u\right |+C\)

Resubstitute \(\displaystyle u\).

\(\displaystyle \frac{1}{2}ln\left | x^2-3x\right |+C\)

Although the integral may look tough, we see a possible relationship between the polynomial of the exponential and the polynomial in front of the exponential.

Let's make our 

\(\displaystyle u=2x^2 + x+ 17\)

\(\displaystyle \frac{du}{dx}=4x+1\)

\(\displaystyle dx=\frac{du}{4x+1}\)

Now let's see the original integral to make the substitutions.

\(\displaystyle \int (8x+2)e^{2x^2+x+17}dx=\int (8x+2)e^{u}\frac{du}{4x+1}\).

At first sight it may seem that nothing cancels out, but at a closer look we can see that the \(\displaystyle (8x+2)\) term can actually be simplified to \(\displaystyle 2(4x+1)\), which can now cancel with the denominator.

\(\displaystyle \int2e^udu\) is our new integral, which simply leads to

\(\displaystyle 2e^u+c\). Now substitute u for what we had earlier.

\(\displaystyle 2e^{2x^2+x+17}+c\)

To evaluate \(\displaystyle \int \frac{x}{\sqrt{x+2}}\: dx\), use U-substitution.

Let \(\displaystyle u=x+2\), which also means \(\displaystyle x=u-2\).  Take the derivative and find \(\displaystyle du\).

\(\displaystyle du=dx\)

Rewrite the integral in terms of \(\displaystyle u\) and \(\displaystyle du\), and separate into two integrals.

\(\displaystyle \int \frac{x}{\sqrt{x+2}}\: dx= \int \frac{u-2}{\sqrt u}\:du=\int u^\frac{1}{2} \:du-2\int u^{-\frac{1}{2}}\:du\)

Evaluate the two integrals.

\(\displaystyle \frac{2}{3}u^{\frac{3}{2}}-4u^{\frac{1}{2}}+C\)

Re-substitute \(\displaystyle u\).

\(\displaystyle \frac{2}{3}(x+2)^{\frac{3}{2}}-4(x+2)^{\frac{1}{2}}+C\)

Pull out the common factor \(\displaystyle (x+2)^{\frac{1}{2}}\).

\(\displaystyle =(x+2)^{\frac{1}{2}}\left[\frac{2}{3}(x+2)-4\right]+C\)

\(\displaystyle =(x+2)^{\frac{1}{2}}\left[\frac{2}{3}x+\frac{4}{3}-\frac{12}{3}\right]+C\)

\(\displaystyle =\sqrt{x+2}\left[\frac{2}{3}x-\frac{8}{3}\right]+C\)

In order to solve this, we must use \(\displaystyle \small u\)-substitution.

Because \(\displaystyle \small \frac{\mathrm{d} }{\mathrm{d} \theta }(cos\theta )=-sin\theta\), we should let \(\displaystyle \small u=cos\theta\) so the \(\displaystyle \small sin\theta\) can cancel out.

We can now change our integral to \(\displaystyle \small \small \small \int_{0}^{\pi /2} \frac{2sin\theta }{\sqrt{u}}d\theta\).

We know that \(\displaystyle \small u=cos\theta\), so \(\displaystyle \small \small \frac{\mathrm{d} u}{\mathrm{d} \theta }=-sin\theta\), which means \(\displaystyle \small d\theta=\frac{du}{-sin\theta }\).

We can substitue that in for \(\displaystyle \small d\theta\) in the integral to get \(\displaystyle \small \small \small \small \int \frac{2sin\theta }{\sqrt{u}}\frac{du}{-sin\theta }\).

The \(\displaystyle \small sin\theta\) can cancel to get \(\displaystyle \small \small \small \small \small \int \frac{-2 }{\sqrt{u}}du\).

The limits of the integral have been left off because the integral is now with respect to \(\displaystyle \small u\), so the limits have changed. This integral can now be solved using the power rule \(\displaystyle \small \small \small \int au^{n}du=\frac{a}{n+1}u^{n+1}+c\) 

which will give you \(\displaystyle \small \small \small -4\sqrt{u}+c\).

Now we can substitute \(\displaystyle \small cos\theta\) back in for \(\displaystyle \small u\) to get \(\displaystyle \small -4\sqrt{cos\theta }+c\). We can bring back our limits now and evaluate because we're in terms of \(\displaystyle \small \theta\) again, which is what our original limits were with respect to.

\(\displaystyle \small \small \small -4\sqrt{cos(\pi/2) }+c\left-\small \small (-4\sqrt{cos(0) }+c\left)=c-(-4+c)=c+4-c=4\)  

TIP: whenever doing \(\displaystyle \small u\)-substitution, don't use the limits until you're in terms of the original variable.

Here, we can use u-substitution. We'll set \(\displaystyle u=1+x^2\) and we'll factor out the \(\displaystyle \frac{1}{2}\) outside the integral.

Now let's calculate \(\displaystyle du.\)

\(\displaystyle du=2xdx\)

And solve for dx.

\(\displaystyle dx=\frac{du}{2x}\).

Plugging these values into the integral we now get

\(\displaystyle \frac{1}{2}\int{\frac{x}{u}\frac{du}{2x}}\).

We now see that the \(\displaystyle x\)'s cancel out and we're left with an integral entirely with u.

\(\displaystyle \frac{1}{4}\int{\frac{1}{u}du}=\frac{1}{4}ln(u) +C\).

We just need to replace u with its original value, doing so results in our final solution.

\(\displaystyle \frac{1}{4}ln(1+x^2)+C\)

To solve the following integral, we must make a substitution to create the following general form:

\(\displaystyle \int \frac{dx}{\sqrt{1-x^2}}=\sin^{-1}(x)+C\)

We make the following subsitution:

\(\displaystyle u=2x, du=2dx\)

The derivative was found using the following rule:

\(\displaystyle \frac{d}{dx}(x^n)=nx^{n-1}\)

The integral now looks like this:

\(\displaystyle \frac{1}{2}\int \frac{du}{\sqrt{1-u^2}}\)

Notice that it is in the same form as the integral we want.

Now use the form from above to integrate:

\(\displaystyle \frac{1}{2}\sin^{-1}(u)+C\)

To finish the problem, replace u with 2x:

\(\displaystyle \frac{1}{2}\sin^{-1}(2x)+C\).

We can simplify

\(\displaystyle \small \int (\ln x)^2 dx\) 

by first doing a substitution, with \(\displaystyle \small u=\ln x\), which gives us \(\displaystyle \small e^u=x\), which means that \(\displaystyle \small e^udu=dx\). So the integral becomes

\(\displaystyle \small \small \int (\ln x)^2 dx=\int u^2e^u du\)

The integral\(\displaystyle \small \small \int u^2 e^u du\) can be solved using two integration by parts, which will give us 

\(\displaystyle \small \small \small \small \int u^2 e^u du=e^u(u^2-2u+2)+C\)

So now we just plg in \(\displaystyle \small \ln x\) into \(\displaystyle \small u\) and \(\displaystyle \small x\) into \(\displaystyle \small e^u\) to get

\(\displaystyle \small \small \small \small \int (\ln x)^2 dx=\int u^2e^u du=e^u(u^2-2u+2)+C=x((\ln x )^2-2\ln x+2)+C\)

Pull out the constant out in front of the integral.

\(\displaystyle \int_{2}^{6}\frac{2}{3xln^3(x)} \:dx=\frac{2}{3}\int_{2}^{6}\frac{1}{xln^3(x)} \:dx\)

Use U-substitution to solve.

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

\(\displaystyle du=\frac{1}{x}\:dx\)

\(\displaystyle \frac{2}{3}\int_{2}^{6}\frac{1}{xln^3(x)} \:dx= \frac{2}{3}\int u^{-3}\:du= \frac{2}{3}\left[ \frac{u^{-2}}{-2}\right]=-\frac{1}{3}\left[\frac{1}{ln^2(u)}\right]_{2}^{6}\)

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

The integral can be solved with a clever substitution:

\(\displaystyle u=\sqrt{2x+3}, du=(2x+3)^{-\frac{1}{2}}dx\)

The derivative was found using the following rules:

\(\displaystyle \frac{d}{dx}(x^n)=nx^{n-1}\), \(\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)\)

Then, when you rewrite the integral in terms of u, you find that you get:

\(\displaystyle \int du = u+C\)

The integration was performed using the following rule:

\(\displaystyle \int dx=x+C\)

Finally, replace the u with our original term.

\(\displaystyle \sqrt{2x+3}+C\)