evaluate the indefinite integral?

Question

Evaluate the indefinite integral
{x + 1} / {x^2 + 2 x} dx

Answer 1

An alternative to splitting it into two integrals or using partial fractions.
∫(x+1)/(x^2+2x) dx

multiply by 2/2.
(2/2)∫(x+1)/(x^2+2x) dx

Use the fact that you can bring constants in and out of the integral and take the top 2 inside but leave the bottom outsi
(1/2)∫[2(x+1)]/(x^2+2x) dx

Distribute the 2
(1/2)∫(2x+2)]/(x^2+2x) dx

let u = x^2+2x
du = (2x+2) dx

Giving you
(1/2)∫du/u

Integrating
(1/2) ln|u| + C
(1/2) ln| x^2+2x |+C

Another similar method that forgoes the initial algebraic manipulation.
∫(x+1)/(x^2+2x) dx
u = x^2 + 2x
du = 2x+2 dx
du = 2(x+1) dx
du/2 = (x+1) dx
1/2∫du/u
1/2 ln|x^2+2x| + C

Answer 2

I get [1/2] [ ln x + ln (x+2)] + constant of integration and the front end simplifies to [1/2] ln (x^2 + 2x)
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The detail(s) you may want to see is/are this / these:

We try representing the fraction {x + 1} / {x^2 + 2 x} as

A/x + B/(x+2) ending up with x+1 = A(x+2) + Bx or (A+B)x + 2A
In order for this to be true for all x, the coefficients of corresponding powers of x must be equal:
(A+B) = 1 and 2A = 1 (solve simultaneously getting A=1/2
and B = 1/2.

Now you're integrating [(1/2) / x + (1/2) / (x+2)] dx which leads to my topmost-line (or answer).

[...and I really should put the logarithm arguments inside absolute value signs.]

Answer 3

?(3x - 2)^20 dx is merely about in the type u^n du u = 3x - 2, du = 3dx. It probable astonishing if I had a three accessible, so i'm merely gonna located on there, nonetheless I even have won to stability it out, so I''ll placed a a million/3 out front. (a million/3)? (3x - 2)^20 3dx now i've got u^du, i will now do u^(n + a million)/(n + a million) (a million/3)(3x - 2)^21 / 21 simplify... (a million/sixty 3)(3x - 2)^21 + C

Answer 4

∫ (x+1)/(x²+2x) dx = ½ ∫ (2x+2)/(x²+2x) dx.
Let u = x² + 2x. Then du = 2x+2 dx
So we get ½* ∫ du/u = ½ ln(u) = ½ ln(x²+2x) + C.

Answer 5

1/2 ln|x^2+2| + C

or maybe easier to see:

ln |x^2+2| + C
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