How do I find the antiderivative?

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Apr 29th 2009, 07:42 AM
anitah

How do I find the antiderivative?

I have a few homework problems, and they're asking me to find the antiderivative. They're quite tough, and I'm really confused. Can anyone be of some help on these few?

integral(4x-5/x^3)dx (/ = over.)

integral 3x over (x^2+4)dx. On this one, the 3x isn't in parentheses.

integral 5xsec^2(x^2)dx

integral (4x-5/x^2)dx

integral e^4x dx

integral sin(x)cos^3(x)dx

Any help on ANY of these is greatly appreciated. I'm really lost (Worried)
Apr 29th 2009, 08:10 AM
Spec

(1) $\frac{3x}{x^2+4}$
We know that $\frac{d}{dx}(\ln(x^2+4))= \frac{2x}{x^2+4}$ , so all we need to do is find a fractional number to multiply/divide $\ln(x^2+4)$ with to get the antiderivative.

(2) $e^{4x}$
We know that $\frac{d}{dx}(e^{4x})=4e^{4x}$ , so you simply need to multiply $e^{4x}$ with $\frac{1}{4}$ to get the antiderivative.

(3) $\sin(x)\cos^3(x)$
$\frac{d}{dx}(\cos^4(x)) = 4\cos^3(x)\cdot -\sin(x)$
Again, look for the appropriate number to multiply/divide with.
Apr 29th 2009, 08:12 AM
Twig

hi

hi

$\int (4x-\frac{5}{x^{3}}) \, dx = 2x^{2} +\frac{5}{2}x^{-2} + C \; ,\, C\in \mathbb{R}$

$\int (\frac{3x}{x^{2}+4}) \, dx = \frac{3}{2}\cdot ln(x^{2}+4) + C$

Some of these are actually quite basic I am afraid.

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