I want to learn exactly how to combine this. If you can show me a step-by-step technique that would certainly be awesome.

You are watching: Integral of ln^2(x)

If you could also point me to some good tutorials top top integration that would be icing on the cake.

What go \$dx\$ mean? I know its delta \$x\$ and very little strips or something (actually I"m not an extremely sure...). If it is for this reason minute (\$ ightarrow 0\$ ?), why isn"t the totality thing simply \$0\$ due to the fact that \$x * 0 = 0\$ where \$x\$ is \$ln(2-x)\$?

Ok, due to the fact that I gained time, I"ll bite, because you obviously absence the yellowcomic.comematical background to understand the remainder of the answers. Identify integration is including up infinitimestally little slices to add up to an area under a curve C. The definition of a identify integral is ;Given a duty that is consistent on the interval \$\$ we division the interval into \$n\$ subintervals that equal broad \$Delta x\$, and from each interval select a point \$x_i\$. Then \$\$int_a^b f(x) yellowcomic.comrmdx = lim_n o infty sum_i=1^n f(x_i) Delta x\$\$

However, the integral you have requested is indefinite, which equals to the antiderivative that \$f\$ which method that after we resolve this, if we differentiate the result we"ll get \$ln(2-x)\$. There are countless integration approaches that you"ll have to learn to handle every integral, yet remember that some attributes have no antiderivative in regards to elementary functions and some don"t have any type of at all!

Let"s begin with u-substitution(which is what you require for this trouble of yours). We"ll try to manipulate the ingredient under the integral sign(which are called the integrand) in order come collapse this integral to an easier one that"s in the table that integrals(or simplify it sufficient to re-manipulate it, but that"s a little more complex, let"s stick to your straightforward problem).

First of all you have to learn this formula;

If friend let \$u=2-x\$ climate you need \$du\$ which will be derived by differentiation \$u\$ wrt. To \$x\$

So allows evaluate \$fracddx u= fracddx (2-x)=fracddx 2 -fracddx x=0-1=-1\$Now if a yellowcomic.comematician look at this he"ll cringe however for simplicity"s sake, let"s treat \$dx\$ algebraically. For this reason you have \$fracdudx=-1 iff du=-dx\$

Let"s go earlier to the integral and insert these new values;

\$\$int ln(2-x) dx= int ln (u)(-du)=int -ln (u) du\$\$

Now we have to learn a new integral building that says \$\$int -f(x) dx=-int f(x) dx\$\$

Let"s usage that new knowledge and get

\$\$int -ln (u) du=-int ln (u) du\$\$

Maybe this is too much, so ns won"t be introducing the integration by parts method to resolve this integral, together it is in integral tables anyway and also we"ll just copy the result. Critical part to keep in mind is the we have actually to include an arbitrary continuous to the result as if we identify that it will go away(remember, we acquire the antiderivative) anyway(but it could have been there, don"t ever before forget it).

Finally we gain this;

\$\$-int ln (u) du=-u (ln(u)-1)+yellowcomic.comrmconstant stackrelu=2-x=-(2-x)(ln(2-x)-1)+yellowcomic.comrmconstant\$\$

It"s every algebra native here.

Now this all might seem a little bit also off or also counter-intuitive, but you"re tho on your way to find out a brand-new branch in yellowcomic.comematics, called calculus, i beg your pardon I find quite beautiful and intriguing. I once started wherein you are and also now my hobby is to shot and solve complicated integrals.

EDIT; Here"s another faster method to resolve your problem if you don"t have any calculus background, and we don"t desire to introduce a lot of new stuff.

Let \$f(x)=2-e^x\$ i beg your pardon is the train station of \$g(x)=ln(2-x)\$. Then \$f(y)=2-e^y\$.

We"ll use the formula \$\$int f^-1(x),dx= x f^-1(x)-Fcirc f^-1(x)+C\$\$ whereby \$F\$ is the antiderivative of \$f\$.

Let"s evaluate \$F(y)\$(using some straightforward integral properties)

\$\$int 2-e^y dy=int 2 dy-int e^y dy= 2y-e^y+c\$\$

Now lets evaluate(using duty and logarithmic properties)

\$\$Fcirc f^-1(x)=2(ln(2-x))-e^ln(2-x)+c=2ln(2-x)-2+x+c=x+2 (ln(2-x)-1) +c\$\$

We have actually all the parts for the formula so we just straight substitute them;

\$\$int ln(2-x) dx=x(ln(2-x))-(x+2 (ln(2-x)-1)))+c=\$\$

Which after algebraic manipulation offers the same an outcome as above.

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This is a special instance where this theorem can simplify points for her case. That usually renders things more daunting so don"t use it on every integral. Pole to common integration techniques.