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How do you find the indefinite integral of #int x^3(x^4+3)^2dx#?

Answer 1

Emilia Aguilar

#(x^4 +3)^2 /12 +C#

Let #x^4 +3 =u#, so that #x^3dx= 1/4 du# On substituting, the given integral becomes #int 1/4 u^2 du#

=#u^3/12 +c#

=#(x^4 +3)^2 /12 +C#

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Answer 2

Paisley Johnson

The answer is #x^12/12+3x^8/4+9x^4/4+c#.

First of all, expand the square: using the formula

#(a+b)^2= a^2 +2ab + b^2#

we have that

#(x^4+3)^2 = x^8 + 6x^4 + 9#

Now we can multiply the expanded square for #x^3#, obtaining

#x^3 * (x^8 + 6x^4 + 9) = x^11 + 6x^7 + 9x^3#

Now this quantity is easy to integrate, because of the linearity of the integral, which means that the integral of the sum is the sum of the integrals:

#\int x^11 + 6x^9 + 9x^3 dx = \int x^11 dx + \int 6x^7 + \int 9x^3 dx#

and each of these integrals can be done using the same rule, i.e.

#\int ax^n dx = {ax^{n+1}}/{n+1}#.

So: #\int x^11= x^12/12#, # \int 6x^7=6x^8/8= 3x^8/4#, and #\int 9x^3 = 9x^4/4#.

The answer is thus #x^12/12+3x^8/4+9x^4/4+c#.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.