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## Calculus 2

# Common integrals review

AP.CALC:

FUN‑6 (EU)

, FUN‑6.C (LO)

, FUN‑6.C.1 (EK)

, FUN‑6.C.2 (EK)

Review the integration rules for all the common function types.

## Polynomials

## Radicals

*Want to learn more about integrating polynomials and radicals? Check out this video.*

*Want to practice integrating polynomials and radicals? Check out these exercises:*

## Trigonometric functions

*Want to learn more about integrating trigonometric functions? Check out this video.*

*Want to practice integrating trigonometric functions? Check out these exercises:*

## Exponential functions

## Integrals that are logarithmic functions

*Want to learn more about integrating exponential functions and start fraction, 1, divided by, x, end fraction? Check out this video.*

*Want to practice integrating exponential functions and start fraction, 1, divided by, x, end fraction? Check out this exercise.*

## Integrals that are inverse trigonometric functions

## Want to join the conversation?

- What is the difference between x^n dx and a^x dx? That is, why is one a polynomial and one an exponential function?(3 votes)
- In the second function, variable x is the exponent. That is why the second one is exponential function.(6 votes)

- Could someone please provide me with the proof for

integral of 1/(a^2 + x^2)(3 votes)- 1/(a² + x²) = 1/(a²(1 + x²/a²)

Let x = a·tan(u)

dx = a·sec²(u) du

Therefore ∫1/(a² + x²) dx = ∫a·sec²(u) / a²(1 + tan²(u)) du = 1/a ∫sec²(u) / (1 + tan²(u)) du

But 1 + tan²(u) = sec²(u)

So ∫1/(a² + x²) dx = 1/a ∫ du = u/a + C

Substituting back for u (= arctan(x/a) ) gives

∫1/(a² + x²) dx = 1/a · arctan(x/a) + C

□(2 votes)

- Why is the integral of tan(x) not listed?(2 votes)
- If you just want to know the answer, then Sal covers it in this video:

https://www.khanacademy.org/math/calculus-home/integration-techniques-calc/reverse-chain-rule-calc/v/integral-of-tan-x

If you feel it's an omission that needs correction (I'd agree) then you might like to raise a "feature request":

https://khanacademy.zendesk.com/hc/en-us/community/topics/200136634-Feature-Requests(3 votes)

- Why isn't there an arccos integral function?(2 votes)
- probably because arcsin and arccos have almost identical derivatives. their derivatives are negatives of each other.(2 votes)

- Where is the video for the second function under "Exponential Functions" (the integral of a^x)? Where are the videos for the whole section of "Integrals that are inverse trigonometric functions" (the integral of 1/sqrt[(a^2)-(x^2)] and the integral of 1/sqrt[(a^2)+(x^2)]? I can't find these three functions mentioned anywhere in the videos.(2 votes)
- I agree some of the things are not explained, I can't find the videos(5 votes)

- All these integrals of trigonometric functions are really confusing for me. Do I have to just learn them by heart? Or is there some section I missed, where they are explained more intuitively?(2 votes)
- There are proofs out there for each trig function but it is much easier to just learn them by heart.(2 votes)

- in common integral review under exponeential functions how integration of ax is ax/ln{a} shoudnt it be like the polynomial example?(1 vote)
- a^x is not a polynomial(4 votes)

- Any mnemonics to remember these? Anyone?(2 votes)
- Just commit the derivatives to memory and then use the opposites to remember these!(1 vote)

- Under the Integrals that are inverse trigonometric functions, why there is "1/a" before arctan but not arcsin?(2 votes)
- Hello,

I one of the practice problems under u substitution: definite integrals gave the problem: ∫ 1/1+9x^2 dx. If you use u substitution you get (1/3)arctan(3x)+c. But if you refer to the integral above it states that ∫1/a^2 +x^2 dx = 1/a arctan (x/a)+c. In my example a=1 but the fraction for the solution isn't 1/1 it's 1/3. I'm confused. If you could please get back to me I'd really appreciate it. Thanks.(2 votes)- When you use u sub to put u in for 3x, you are also subbing in du/3 for dx, let me know if that doesn't make sense. But it should look like this now ∫1/1+u^2 du/3 Anyway, now with du/3 or in other words (1/3)du, you can bring that 1/3 out to the front, and make it look like (1/3)∫1/1+u^2 du and solve ∫ 1/1+u^2 du = arctan(u/1)+c = arctan(3x/1)+c, but you still have that 1/3 out in front. So it's (1/3)arctan(3x)+c. Again, let me know if that doesn't make sense.(1 vote)