Question

Indefinite Integral Calculation: S=integral

S(X^2+4x)/(x^3+6x^2+5)dx

So I get that I have to do the u substitution:
u = x^3 + 6x^2 +5 --> du = 3x^2 +12x
--> 1/3du = x^2 +4x dx

then
--> 1/3 S 1/u du
--> 1/3ln(u) --> then i plug u back in.

The problem I am having is the symbolab calculator that I'm using to check my work is adding this extra step that I don't understand.

it gets u = x^3 +6x^2 +5
then it gets du = 3x(x+4)
then dx = 1/3x(x+4) du

it plugs it in as:
S (x^2+4x)/u (times) 1/(3x(x+4)du
then 1/3u du, after which it lines up with what I got.

help pl0x :)

Answer 1

u get\(\int \frac{1}{3u}du\) ?

Answer 2

yeah

Answer 3

I just don't understand why they solved for dx after getting du, cant you just plug du back in to get what you need?

Answer 4

i really don't understand what is confusing, both methods are equal,
the calculator has just added one extra step,

Answer 5

@baru I don't understand the one step thought, that's what I want to understand.

@robtobey good one lol. Doesn't answer my question though

Answer 6

\(u=x^3+6x^2+5\\du=(3x^2+12x)dx\\du=3(x^2+4x)dx\\du=3x(x+4)dx\\dx=\frac{du}{3x(x+4)} \)

Answer 7

\(\int \frac{x^2+4x}{x^3+6x^2=5} dx\\\\rearrange~numerator\\\int \frac{x(x+4)}{x^3+6x^2=5} dx\)

Answer 8

substitute for for u and dx
\(\int \frac{x(x+4)}{u} \frac{du}{3x(x+4)}\)

Answer 9

x(x+4) in the numerator and denominator get cancelled
leaving \(\int \frac{du}{u\times 3}\)

Answer 10

@baru, it's true what they say..

you really are a human calculator. lol thanks