The integral we were given is improper, because the upper bound of integration when plugged into the integrand gives us  which is equal to infinity. 

To integrate, we must do the following:

.

We evaluate the limit from the left because the upper limit of integration was improper. 

When we integrate, keeping the limit outside of the integral, we get

This is a standard definite integration in which the upper limit of integration and lower limit of integration are plugged into our result from the integration, and the resulting functions are subtracted from each other. 

The integral was performed using the following rule:

Finally, we evaluate the limit:

As the natural log approaches zero from the left (positive), the function approaches negative infinity.

The integral is improper because the upper limit of integration is infinity. To integrate, we must do the following:

We evaluate the limit from the left because t replaced the upper limit of integration.

Next, keeping the limit on the outside of the integral, integrate and evaluate the limit:

The integration was performed using the following rule:

Because this was a definite integral, we plug in the upper limit and lower limit of integration into the output function and subtract as shown above.

When evaluating the limit, the dominant term is infinity being squared, cubed, and raised to the fourth power, which is enormously big!

The integral is improper because of the upper limit of integration (creates zero in the denominator of the function being integrated). So, we have to do the following:

We evaluate the limit from the left side because the upper limit of integration was the one that caused problems.

Now, to integrate we must do a substitution, but this also means changing the limit:

The derivative was perfomed using the following rule:

Now, rearrange, and rewrite the limit in terms of our new t value, which originally was :

Next, perform the definite integration, keeping the limit:

The integration was performed using the following rule:

Now, when we evaluate the limit, we notice that the natural log function, as it approaches zero, approaches negative infinity. The fact that there is a coefficient on the limit or the additional term being subtracted are insignificant compared to the negative infinity term. So, our final answer is .

Because this is an improper integral, you must rewrite it as so: 

.
You can now solve the integral by substuting  and  to recreate the integral as  which then becomes after integration  evaluated at t and 0.

Substituting arctanx back in for u and then evaluating at the bounds gives you the following: 

.

The arc tangent of 0 is just 0. But remember that t is approaching infinity. Below is the graph of arctan. 

As you can see, as x (or in this case t) approaches infinity, the graph slowly approaches its horizontal asymptote at pi/2. And so,  can be seen to be .

So the answer to the problem is 

 which simplifies to 

The first thing we have to do is set up the integral again in a workable form:

Now we need to start evaluating our new integral. The best way to go about this is to use trig substitution. 

set 

Now we must solve 

and substitute back in later. Recall that .

then our integral is equal to 

which, from basic trig identities, we know comes out to 

Now we are ready to resubstitute the tangent function back in. We can again use trig identities, and notice that

thus . Now we replace tan  with  and get that

The next step is to find the limit:

The first term clearly goes to 0 because the bottom has a higher degree than the top, and the inverse tangent has a known limit as a goes to infinity: pi/2. So our final answer is

When evaluating an improper integral, we substitue infinity for a constant, and then evaluate the integral as the constant goes towards infinity, which looks like this:

, let "t" = infinity, which gives a new integral: 

evaluating the integral, we get: 

Since equals 0, and  is a constant, unaffected by what happens to "t", our integral is ,which equals

The integral we are given is improper at the lower bound (we have a zero in the denominator at the lower bound), thus we need to use the following limit to evaluate the integral:

The integral was evaluated using the following rule:

, with the following subsitution made: 

(The derivative was found using the following rule:

.)

Now, we must evaluate the result at the upper and lower bounds:

As the natural logarithm goes to zero, the output is negative infinity. The negative sign in front means that our answer is . (Note that infinity dominates the constant it is being added to.)

The integral we are asked to solve is improper because its upper bound is infinity. To remedy this, we must rewrite the integral as the following:

Next, we must make the following substitution:

Now, we can rewrite the integral in terms of u:

Note that the constant was moved outside of the limit, and the limits of the integration were kept in terms of x.

Next, we integrate:

The integral was performed using the following rule:

Note that we replaced u with our original term and removed the absolute value sign because the argument is always positive.

Now, we perform the definite integration, and we get

As the natural logarithm goes to infinity, its output equals infinity. 

Let , and  

Since , 

By the identity of  :

By the fundamental theorem of calculus: 

By properties of logarithms:

By the power rule:

By the fundamental theorem of calculus: