Notes: it's important to choose

$$ x = u \rightarrow dx = du$$

because by doing so we're reducing the monomials degree (from 1 to 0). If we choose

$$ x = dv \rightarrow v = \frac{x^2}{2} $$

we increase the degree (from 1 to 2) and we complicate the integral more because the exponential factor remains the same and we're left with this integral

$$ \int {\frac{x^2}{2}\cdot e^x }dx$$

Notes: As seen in the exercise before, it doesn't matter if cos(x) in u or dv (due to the fact we obtain a sinus). We choose u = x to reduce it's degree (and that way x disappears). If we choose dv = x, we increase the degree.

$$ dv = x \rightarrow v = \frac{x^2}{2}$$

In this integral we don't have an explicit product of functions, but we don't know what the logarithms primitive function is, so we differentiate it, that way u = ln(x).

It's in our interest to select u = x² (to reduce the exponent) but then we're forced that dv = ln(x) and obtaining v isn't immediate. So we'll select the other case

If we chose dv = ln(x), we won't be able to obtain v easily. It's better to chose u = ln(x)

Usually, we would select u = x² to reduce the exponent, but then we would end up with dv = arctan(x) and we don't know the primitive of arctan. So we will select

Now we have to calculate the integral of a rational function. In order to simplify the expression we're going to divide the polynomials:

$$\frac{P(x)}{Q(x)} \rightarrow P(x) = Q(x)C(x) + R(x) $$

where C(x) is the quotient and R(x) is the remainder.

Dividing the expression by Q(x) we get

$$ \frac{P(x)}{Q(x)} = C(x)+ \frac{R(x)}{Q(x)}$$

We'll use this break down in the integral:

We resolve the integral:

Then

Note: We have taken the absolute value away from the logarithm because it always has a positive argument.

Every time we integrate or differentiate cos(x) we obtain ±sin(x). So, it doesn't matter if it's u or dv. However, it's better to choose u = x² because when we differentiate we reduce the exponent: du = 2x. We will select dv = cos(x).

We integrate by parts again, but we have to chose u = x, if not we return to the previous step:

Then,

We choose u = x to reduce its exponent (and that way x disappears).

The primitive of

$$ \frac{1}{cos^2(x)}$$

is immediate:

Similar to what happens with sin(x) and cos(x), when we differentiate or integrate e^x we obtain e^x, so it's indifferent whether it's u or dv. If we select the exponential to be u, this factor will always stay in the integral and furthermore the monomial (power) will be dv, and we will increase its degree when calculating v. So we'll select dv = e^x and the monomials of the polynomial as u to reduce the exponent until it's a constant.

In this example it doesn't matter which factors are u and dv, because when integrating and differentiating e ^-x we obtain –e ^-x and when integrating and differentiating cos(x) we get ±sin(x). This is a cyclical integral in which we have to apply integration by parts twice (with the same choices so we don't go backwards) and we have to isolate the integral from the mathematical expression we obtain.

We have a exponential multiplied by sine once again, so we are up against a cyclic integration because we must apply the integration by parts process twice(always with the same selection so we don't go back a step) and isolate the integral from the expression we obtain.

We can chose u and dv as we please.

We choose the polynomial as u to reduce the exponents until they disappear.

Each time we differentiate or integrate exponentials we obtain the same exponential but multiplied by a constant (or the inverse of said constant), so which is u or dv doesn't matter. We select in function of the other factor and seeing it's a monomial we make u = x² to reduce the exponent:

The integral of arcsine can be considered direct, but we can also calculate its primitive using integration by parts:

This is a cyclical integral in which we have to apply integration by parts twice (with the same choices):

We have to apply integration by parts three times (with the same choices):