Question

Hello, I need some Calculas help...

Answer 1

I was wondering if you check my answer... all help is appreciated...

Answer 2

At first glance that solution does look reasonable. You'd need the u-substitution with the denominator, so the natural log makes sense. The 1/4 factor because the du = 4 .. dx, so 1/4 du. That looks good to me I think.

Answer 3

could u guys help check my answers for a few more

Answer 4

The u-sub on 3x^3 because x^2 in the integrand. u=3x^3, du=9x^2 dx, 1/9 du = x^2 dx and so integrate 1/9 e^(u) du = 1/9 e^u + c = 1/9 e^(3x^3)

If these are all indefinite integrals, you can also take the derivative to check the answer. The derivative just returns back to the integrand, like
d/dx ( 1/9 e^(3x^3) ) = 1/9 * d/dx(3x^3) e^(3x^3) = 9/9 x^2 e^(3x^3)

Answer 5

ok thnxs

Answer 6

Glad to help! :)

Answer 7

there were 2 I did not know how to solve at all, Ill post them, btw thnxs for your help and teaching me every step and such

Answer 8

actaully one more haha

Answer 9

For each of those first two, it may help things out if you can get an exponential base of e.
\(a^ x = \left( e^{\ln a} \right) ^x = e^{x \ln a}\)
At the end you can convert back to base a by the reverse process. Does that make sense?

Answer 10

ok

Answer 11

Alternatively, if you know this as an identity from derivative calculus:
\( \dfrac{d}{dx} a^x = a^x \ \ln a \)
The reverse process is just: \( \displaystyle \int a^x \ dx = \dfrac{a^x}{\ln a} \) We just need u-substitution to deal with the exponent of -2x.

You can eliminate some choices automatically, because all exponentials act like e^x. Their derivatives are a constant multiple of themselves. It isn't possible to go from 5^(-2x) to 5^x.

Answer 12

ok thnxs

Answer 13

If that clears up confusion for the first two, the last was that number 8.
The derivative of tan x is sec^2 x. That is the same as 1/cos^2 x, right? So we have tan x and its derivative sec^2 x.

Answer 14

yup

Answer 15

So if we u-substitute u = tan x, du = sec^2 x dx.= 1/cos^2 x dx. Can you see how to work out the problem with that out of the way? (It is about the same problem as in integrating 5^(-2x) dx at that point).