Integration by Parts Homework Help - Q2

• Goldenwind
In summary, the conversation was about a student preparing for an exam and solving end-of-chapter questions. They encountered some difficulties with integrals and asked for help on specific problems. The conversation covered techniques such as substitution and the chain rule, as well as the general rule \int \frac{f'(x)}{f(x)}dx = ln|f(x)| + C. The conversation also discussed the integral of arcsin(x).
Goldenwind

Homework Statement

In preparation for an exam next week, I'm solving some of the end-of-chapter questions. There are 30 questions. I've solved a few on my own, but here's one I'm getting stuck on.

Question 2)
$$\int x5^x dx$$, problem here is I don't know how to deal with $5^x$.
The correct answer should be: $$\frac{x}{ln5}5^x - \frac{1}{(ln5)^2}5^x + C$$

To avoid spamming the forum with many threads, I'll just post in this one whenever I hit an integral I'm unsure of. For now, it's just the one above.

Homework Equations

$$\int u*dv = uv - \int v*du$$

Last edited:
It might look a little easier if you write 5^x=exp(ln(5)*x). Put the exp into the dv part.

Did another one here:

Question 4)
$$\int x log_{10}x dx = \int x \frac{ln(x)}{ln(10)} dx$$
$$= \frac{1}{ln(10)}\int xln(x) dx$$
$$= \frac{1}{ln(10)}\int D(\frac{x^2}{2})ln(x) dx$$
$$= \frac{1}{ln(10)}\left(\frac{x^2}{2} * ln(x) - \int \frac{x^2}{2} D(ln(x))dx\right)$$
$$= \frac{1}{ln(10)}\left(\frac{x^2}{2} * ln(x) - \int \frac{x^2}{2} * \frac{1}{x}dx\right)$$
And yeah, getting too lazy to TeX it all. Question is, when I finally get rid of all integrals, the +C will be inside the 1/ln(10) bracket. Do I write C/ln(10), or do I assume that whatever C is will already have that applied, so I just write C?

Yes, the constant C/ln(10) is the same as the constant C' - and it isn't even necessary to distinguish between "C" and "C'".

Dick said:
It might look a little easier if you write 5^x=exp(ln(5)*x). Put the exp into the dv part.

So, 5^x = exp(ln(5^x)) = exp(xln(5)) = e^(xln(5))... I see that much.
I also know the derivative/antiderivative of e^x = e^x.
What is the derivative/antiderivative of e^(xln(5))??

Use the chain rule. The derivative of e^(a*x)=a*e^(a*x). What's the antiderivative?

e^(ax) / a, methinks... 'cause then taking the derivative of that would pull down another 'a', canceling out the division. Makes sense.

Thanks :)

Why is $$\int \frac{D(1+x^2)}{1+x^2}dx = ln(1+x^2)$$?

I know $D(ln(x)) = 1/x$, however I don't see how my book did this calculation.

The point is that $\int\frac{du}{u}= ln|u|+ C$, No matrer what "u" is. In this case u= 1+ x2.

It is from the more general rule that

$$\int \frac{f'(x)}{f(x)}dx = ln|f(x)| + C$$

which comes from the chain rule. If you think about substituting u for 1 + x2, then what is du in terms of dx?

If u = 1 + x^2, then du = 2x.
Thanks for the ln|blah| tip! I remember being taught it, but had forgotten.

One final question for tonight.
I'm in the middle of integrating arcsin(x), and as a segment, I need to compute the antiderivative:
$$\int \frac{x}{\sqrt{1 - x^2}}$$

From looking up the answer, I see that this integral is equivalent to $\sqrt{1 - x^2}$. Why would this be? I see how it can be broken up into $\frac{x}{1} * (1 - x^2)^{-\frac{1}{2}}$. Does this help me, or am I missing another rule?

Yes, you are right, although the division by 1 is not really necessary. Think chain rule again, and see if you can separate one part of the integral as a derivative of another part, then substitute.

Derivative of the top = 1

Derivative of the bottom = $$-2x * (1 - x^2)^{-\frac{3}{2}}*-\frac{1}{2}$$
= $$x(1 - x^2)^{-\frac{3}{2}}$$

Sorry, but I still don't see it >.<;
I see how the chain rule was used to find the derivative of the bottom, but otherwise...

What is integration by parts?

Integration by parts is a method used to integrate a product of two functions by breaking it down into simpler parts. It involves using the product rule from differentiation and the fundamental theorem of calculus to simplify the integral.

When should integration by parts be used?

Integration by parts is typically used when the integral involves a product of two functions where one function is easily integrable and the other is not. It is also useful when the integral involves a combination of polynomials, logarithmic, or exponential functions.

What are the steps for integration by parts?

The steps for integration by parts are as follows: 1) Identify the two functions in the integral, one will be considered the "u" function and the other the "dv" function. 2) Use the product rule to rewrite the integral as the product of the two functions. 3) Integrate the "dv" function and differentiate the "u" function. 4) Substitute the values into the integration by parts formula. 5) Simplify and solve for the integral.

Can integration by parts be used for definite integrals?

Yes, integration by parts can be used for both indefinite and definite integrals. For definite integrals, the limits of integration must be substituted into the final answer.

What are some common mistakes when using integration by parts?

Some common mistakes when using integration by parts include choosing the wrong "u" and "dv" functions, forgetting to apply the product rule, and making errors in the substitution step. It is important to carefully choose the functions and double check the work to avoid these mistakes.

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