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Complicated Indefinite Integrals - Problem 1 3,968 views
Some integrals require a knowledge of specific antiderivatives. For example, recall that the derivative of ln(x) is 1/x. So, the antiderivative of 1/x is ln|x| + c. This means that when 1/x or x-1 is in an integral, remember that you cannot use the power rule on it, and that its antiderivative is ln|x| + c. Then, you can antidifferentiate as you normally would.
Let’s solve some harder antiderivative problems. I’m asked to perform the antidifferentiation integral of 4x to the -1 plus 10x to the -3. And what I have here are just two powers of x. So all I really need to do is, separate these two terms and then pull the constants out and then I can integrate them directly using this formula up here. This is my power rule for antidifferentiation. Let me do that.
I’m going to do both steps in one. I’m going to separate the integral of the sum into the sum of integrals and pull the constants out, in one step. 4 times the integral of x to the -1dx plus, 10 times the integral of x to the 3dx. Now you’ll notice I have x to the -1 and x to the -3. This is a very special power of x, because the top part of the formula of the power rule, doesn’t apply to x to the -1. This doesn’t work if the power is -1, you get division by 0.
You use this formula instead. Very important. That’s going to be 4 times natural log of the absolute value of x, plus 10 times, and then antidifferentiating this, I can use the top rule. I add one to the exponent, -3 plus 1, is -2, and I divide by the same number -2. Then I just add a c at the end. So my antiderivative is going to be 4 ln of the absolute value of x plus 10 over -2, is going to be -5. So let me just make that minus 5x to the -2 plus c. You can also write that 4ln of the absolute value of x minus 5 over x² plus c. Those are my antiderivatives of 4x to the negative 1 plus 10x to the -3.
Let’s take a look at another example. You have to be careful when you’re integrating something like this. Some binomial raised to a power, sometimes it’s best just to multiply this out and it is in this case. Let me do that. (1 over 2x)² is going to be (1 over 4x)² and then at the very end I’m going to get a plus 6², so 36. The middle term is going to be twice the next product. So 2 times 1/2x times 6, -2 times 1 over 2x times 6. That’s going to be -1 over x times 6, so -6 over x, that’s the middle term. Now, these are basically just powers of x. I need to write them as such. So when integrate, when I treat these as ¼ x to the -2, minus 6x to the -1 plus 36.
Now I can separate the integral. The integral of this sum is the sum of integrals. This becomes ¼ times the integral of x to the -2dx minus 6 times the integral of x to the -1dx plus the integral of 36.
For this term, I’m going to use the power rule. N is not -1, so I can just add one to the exponents. -2 plus 1 is -1, so it’s ¼ x to the -1 over -1, minus 6. Now I integrate x to the -1, I do get natural log of absolute value of x. Plus, the integral of a constant is that constant times x. So this will be 36x and I throw a plus c in the end. Just about done. I have -1/4 times 1 over x, that’s -1/4x minus 6ln absolute value of x plus 36x plus c.
That’s my answer. So remember when you’re dealing with an expression like this that involves multiplication of powers, sometimes it's best just to multiply that out and convert everything into the powers of x. You can use your power rule of antidifferentiation, which is what we did here and here. And just remember that x to the -1 integrates differently than x to any other power.