Solving Indefinite Integrals: Tips, Hints & Help

[rooski]

I am having much trouble with indefinite integrals - i get most of the basic theory behind them but as soon as i am confronted with a larger more complex question i get stuck too easily.

These questions are not for my homework, they are just practice for my test. Any hints, tips and general help is appreciated.Homework Statement

1. $$\int x^{2} / (x^{2} - 4) dx$$

2. $$\int (x + 1) ln x dx$$

3. $$\int (2 - \sqrt{x})^{2} / x dx$$

4. $$\int sec^{2} x \sqrt{1 + tan x} dx$$

5. $$\int cos^{2} x sin^{3} x dx$$

Attempts.

1. I started with u substitution and made u = x^2. Since du = 2x, i did $$\int x / ( x^{2} - 4 ) x dx$$ is this proper?

2. Integration by parts... $$\int ( x + 1 ) ln x dx = ( 1/2x^{2}ln x ) + ln x^{2} - 1/4 x^{2} + x + C$$ - is this right?

3.

4. I know that $$\int sec^{2} x = tan x$$ but that's the extent of my progress.

5.

any help appreciated.

i will be posting more problems and attempts as i continue to get stumped..

 

[Char. Limit]

For 1, I recommend rewriting the numerator as (x^2-4)+4. Then work with that.

For 2, expanding then taking care of each integral individually will help.

For 3, expansion again will save you.

 

[rooski]

For #1 i assume you mean for me to do that so i can make u = x^2 - 4... Right? If that is the case then i need to find a way to incorporate du/2 = x into it though... Since du = 2x.

For #2 i also cannot find where 1/x dx fits into it. I made u = ln x.

Also i have a new question... $$\int sin^{5} x dx$$ - not sure what to do in the event of an odd power.

 

[Char. Limit]

Well, for #1, by making the numerator x^2-4+4, you can rearrange your integral to this:

$$\int \frac{x^2-4+4}{x^2-4} dx = \int 1 + \frac{4}{x^2-4} dx$$

The second integral in #1, if I have it right, requires a trig substitution.

For #2, you have x ln(x) + ln(x). Make that two integrals...

$$\int x ln(x) dx + \int ln(x) dx$$

And then solve each integral individually by parts.

For your new one, change sin^2(x) = 1 - cos^2(x). Then you have $$\int sin^3(x) - sin^3cos^2(x) dx$$. Repeat to fix the sin^3 term.