Find the antiderivatives.
1) �ç�ãx (x+1) dx
2) �ç(x^3-4)/ (x^2) dx
3) �ç(3sinx) dx
4) �ç(5secxtanx) dx
5) �ç(1/x) dx
6) �ç(e^x - 2x) dx
7) �ç(3)/((1-x^2)^1/2) dx
I picked the antiderivatives i did not know how to do. Please show me how to do a couple of these.
the weird symbols in front are supposed be integral signs but watever. i got number 1 and i have the rest i need someone to check the rest.
2)(1/2x^2)/(-4x)
3)-cosx
4)5secx + c
5)0
6)e^x - x^2
7) 3sinx
#2, I would change
(x^3-4)/ (x^2) to x - 4x^-2
and the integral of that is
(1/2)x^2 + 4x^-1 + c
check by differentiating my answer, it is right.
#3, you had
-cosx, how could you get the 3sinx from that?
should have been -3cosx + c
#4
if y = secx
isn't y' = secxtanx ??
take care of the 5, and you are done
#5, you should recognize some standard functions and their derivatives,
if y = lnx
isn't y' = 1/x ??? , so .....
#6, good
#7, how can the derivative of 3sinx, your answer, possibly be (3)/((1-x^2)^1/2) ?????
Sure, I can help you with a couple of these antiderivatives. Let's start with the first one:
1) To find the antiderivative of x(x+1) dx, we can apply the power rule and the constant rule of integration.
Using the power rule, we raise the power of x by 1 and divide by the new power:
∫x(x+1) dx = ∫(x^2 + x) dx = (1/3) * (x^3) + (1/2) * (x^2) + C
Here, C is the constant of integration.
Let's move on to the next one:
3) To find the antiderivative of 3sin(x) dx, we can use the trigonometric identity:
∫sin(x) dx = -cos(x) + C
Applying the constant multiple rule of integration, we have:
∫3sin(x) dx = 3 * (∫sin(x) dx) = 3 * (-cos(x)) + C = -3cos(x) + C
Now you can try solving the others using similar techniques. Remember to apply the appropriate rules of integration and simplify as much as possible.