The standard integrals, composites with ax + b, and spotting when the inside derivative is already there.
Pick an integral. The left readout is the thing being integrated; the right is the proposed answer differentiated back. If the answer is right they agree at every x, so move the slider and try to break it.
∫ 2x(x² + 1)⁴ dx
= (x² + 1)⁵⁄5 + C
| ∫ f(x) dx | xn | sin x | cos x | ex | 1x |
|---|---|---|---|---|---|
| gives | xn+1n + 1 | −cos x | sin x | ex | ln|x| |
n = −1 is allowed here. At 5.5 the index had to be a whole number other than −1, so 1⁄x was out of reach. Now it integrates to ln|x|, and that is why anything of the form 1⁄(ax + b) becomes a logarithm rather than a power. Reaching for "add one to the power" on 1⁄x gives division by zero.
If the inside is ax + b, its derivative is just the number a, so you integrate the outside and divide by a.
∫ k g′(x) f(g(x)) dx works when the inside derivative is already present as a factor. A constant you can adjust by dividing. A missing variable you cannot conjure.
∫ sin(x²) dx has no elementary answer, and you are not asked for one. If the inside is not linear and its derivative is not sitting there, the integral is not one you are expected to do.
1. ∫ 1x dx is:
2. ∫ cos(2x + 3) dx is:
3. Which of these can be integrated by inspection?
Naming the inside function. Writing u = x² + 1 and du = 2x dx is never wasted, even when you can see the answer.
The dividing constant. Differentiating your answer back takes five seconds and catches nearly every error on this sub-topic.
Keep the modulus in ln|…| and keep the + C. Both are routinely dropped and both are sometimes a mark.
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