help with integration for calculus ?

thats correct. Here hows its done

x^2+3 div x^3+1

x would go into it

so:

x^3+0x+1

x^3+3x+0

gives -3x+1 as a remainder

so its x+ (-3x+1)/(x^2+3)

now you can integrate just the x and well keep that aside and work with the -3x+1/x^2+3

now we have the following (1/x^2+3 - 3x/x^2+3)dx

we can separate these two

int 1/x^2+3 dx - int 3x/x^2+3 dx

lets do int 3x/x^2+3 dx first

u=x^2+3

du=2xdx

dx=du/2x

so int 3x/u*du/2x = 3/2u*du

3/2 ln(x^2+3)

now for

int 1/x^2+3 dx

general formula for these are 1/x^2+a^2 = 1/a arctan(x/a) +c

so we can put this in this as

1/x^2+(sqrt(3))^2 dx

now we have 1/sqrt(3) arctan (x/sqrt(3))

put it all together and we get

(1/sqrt(3) arctan (x/sqrt(3))) - (3/2 ln(x^2+3))

Edit bah forgot the first int x dx... x^2/2

x^2/2 + (1/sqrt(3) arctan (x/sqrt(3))) - (3/2 ln(x^2+3))

there you go.

The key to all this was learning the inverse trig integrals:

Namely 1/x^2+a^2 = 1/a arctan(x/a) +c

I gather your at that section. memorize it and it will save you a lot of headaches.

You started off correctly. When you divide the two polynomial you get a quotient of x and a remainder of -3x+1, so your new integral is:

x-(3x+1)/(x^2+3)

I'll assume you can integrate x, so we can move on with the second term, (3x+1)/(x^2+3). This one you'll have to split into the sum of:

3x/(x^2+3) + 1/(x^2+3) - we'll deal with the former first.

The derivative of the denominator is 2x, so we'll have to multiply and divide the polynomial fraction by 2/3, leaving:

(3/2)* Integral of[2x/x^2+3]. Integrate this to get (3/2)Ln(x^2+3)

For 1/(x^2+3) we have to notice that it somewhat resembles the arctan integral, which is (arctanx)'=1/(x^2+1).

We will have to divide top and bottom by 3 to get rid of that 3:

(1/3)* integral of[1/(x/sqrt3)^2+1]

If you integrate that it will give you:

(1/3)arctan(x/sqrt3)

Putting the 3 integrated terms together you get your answer.

Division yields:

x + (1 - 3x)/(x^2 + 3)

The x you can integrate. As for the second part, it can be broken up as follows:

(1 - 3x)/(x^2 + 3) = [1/(x^2 + 3)] - [(3x)/(x^2 + 3)]

Does that help?

∫ (x^3 +1)/(x^2 +3) dx

split this as

∫ x^3 dx / (x^2+3) + ∫ dx /(x^2+3)

Consider

∫ x^3 dx / (x^2+3) = ∫ x^2 x dx / (x^2+3)

Let u=x^2

du=2x dx

x dx = (1/2) du

The first integral becomes

(1/2) ∫ u du / (u+3)

(1/2) [ ∫ (u+3-3) / (u+3) du ]

=(1/2) ∫ du - (3/2) ∫ du/(3+u)

=(1/2) u - (3/2) ln (3+u)

=(1/2) x^2 - (3/2) ln (3+x^2) --------(1)

Now, consider the second part of the integral

∫ dx /(x^2+3) = ∫ dx /(x^2+(√3)^2)

You can recognize this as an arctan integral

= (1/ √3) arctan( x/ √3) ----- (2)

Adding (1) & (2)

(1/2) x^2 - (3/2) ln (3+x^2) +(1/ √3) arctan( x/ √3)+C

is the answer

(x^3 +1)/(x^2 +3) = x^5 + 3x^3 + x^2 + 3

integral(x^5 + 3x^3 + x^2 + 3) = 1/6x^6 +3/4x^4 + 1/3x^3 + 3x + constant

(x³+1)(x²+3) = x^5 + x² + 3x³ + 3

Integral: ((x^6)/6)+((3x^4)/4)+((x^3)/3) + 3x + c

Sorry about the dodgy typing, it won't let me go above x³, and for the second term, it can either be 3x to the 4 all divided by 4, or 3/4 x to the 4. Hope my rambling makes sense, it doesn't to me...