(1)

\[

\begin{aligned}

\int \frac{x}{x^2+2x-3}dx= \int \frac{x-1+1}{(x+3) (x-1)}DX

= \int \frac{1}{x+3} + \frac14 \frac{4}{(x+3) (x-1)} DX

\\=\int \frac{1}{x+3} + \frac14 \frac{(x+3)-(X-1)} {(x+3) (x-1)} DX

= \int \frac{3/4}{x+3}-+ \frac{1/4}{x-1} DX

= \frac34 \ln|x+3| +\frac14 \ln|x-1|+c.

\end{aligned}

\]

In fact, the formula does not feel the convenience of a determined coefficient, such as

\[

\frac{x}{(x+3) (x-1)} = \frac{a}{x+3} + \frac{b}{x-1},

\]

Multiply the left and right sides multiplied by $x +3$, make $x =-3$ $A =\frac34$, and both sides multiply $x -1$, again make $x =1$, get $B =\frac14$. Higher efficiency, so the basic method must be mastered.

(2) because

\[

(x^2+2x+3) ' = 2x+2

\]

So

\[

\begin{aligned}

\int \frac{x+2}{x^2+2x+3}dx=\frac12 \int \frac{(x^2+2x+3) ' +2}{x^2+2x+3}dx=\frac12 \ln|x^2+2x+3| +\int \frac{1}{(x+1) ^2+2}DX

\\= \FRAC12 \ln|x^2+2x+3| +\frac{1}{\sqrt 2} \arctan \frac{x+1}{\sqrt 2}+c.

\end{aligned}

\]

(3)

Because

\[

\frac{x^2+1}{x (x-1) ^2} = \frac{1}{x} + \frac 2 {(x-1) ^2}

\]

So

\[

\int \frac{x^2+1}{x (x-1) ^2}dx =\ln |x| -\frac{2}{x-1}+c.

\]

(

Description: Make

\[

\frac{x^2+1}{x (x-1) ^2} = \frac{a}{x} + \frac B {(x-1) ^2} + \frac C{x-1}

\]

Multiply the left and right sides by $x $, and then make $x =0$ $A =1$, on both sides of the same multiplied by $ (x-1) ^2$, then make $x =1$ $B =2$, the last two sides make $x =-1$ by $A, b$ worth $C =0$.

)

(4)

\[

\int \frac{3}{x^3+1}dx =\int (\frac{1}{x+1} + \frac{-x+2}{x^2-x+1}) dx

= \ln|x+1| -\FRAC12 \ln|x^2-x+1| + \sqrt 3 \arctan (\frac{2\sqrt 3}{3} (X-1/2)) +c.

\]

(5) Method one: For the classical transformation, that is, the elimination of $1+x^2$ items, so that $x =\tan T $, then

\[

\begin{aligned}

\int \frac{1}{(1+x^2) ^2} dx= \int \cos^2 t dt = \frac \int (1+\cos 2t) dt = \frac T+\frac14 \sin 2t +C=\FRAC12 T +\f RAC \sin T\cos T +c,

\end{aligned}

\]

So

\[

\int \frac{1}{(1+x^2) ^2} dx = \frac12 \arctan x + \frac12 \frac{x}{1+x^2}+c.

\]

Method Two: (If you remember the book example recursive formula, can be used directly, otherwise, the basic idea is that we do not come out of the reason is the denominator is too high, you can choose two, one is inverted transform, a book on the way to do the division)

\[

\begin{aligned}

\int \frac{1}{(1+x^2) ^2} dx =-\frac 12\int \frac 1{x} d (\frac 1{1+x^2})

=-\frac12 \frac 1{x (1+x^2)}-\frac12 \int \frac 1{x^2 (1+x^2)} DX

\\=-\FRAC12 \frac 1{x (1+x^2)} + \frac 1{2x} +\frac12 \arctan X+c

= \frac12 \frac x{1+x^2} +\frac12 \arctan x+c.

\end{aligned}

\]

Method Three: Make the $t =\frac 1x$, then

\[

\int \frac{1}{(1+x^2) ^2} dx =-\int \frac {t^2}{(1+t^2) ^2} DT =\frac12 \int t D (\frac 1{1+t^2}) =\frac12 \frac{t}{1+t^ 2}-\FRAC12 \int \frac{1}{1+t^2} DT

= \frac12 \frac{t}{1+t^2}-\frac12 \arctan t+c.

\]

The results are all the same.

(Note: It is also possible to see an application of partial integrals, partial integrals are generally used to eliminate logarithmic functions and inverse trigonometric function, and another function is if there are complex items, and high-level words, such as here $\frac 1{(1+t^2) ^2}$, you can consider using a partial integral, Make it simple, of course, these are not omnipotent, because of the topic and different. So the key is the first triangle transformation, is a kind of unified mechanized method. )

(6) Order

\[

T= \sqrt{\frac{x}{2-x}}, \qquad \mbox{} x= \frac{2 t^2}{1+t^2}, \qquad dx = \frac{4t}{(1+t^2) ^2}dt

\]

So

\[

\int \frac{1}{x^3} \sqrt{\frac{x}{2-x}} D x =\int \frac{1+t^2}{2 t^4} DT

\]

Slightly lower

(7) Make $t = (x-1) ^{1/6}$, then

\[

\int \frac{1-\sqrt{x-1}}{1+\sqrt[3]{x-1}} DX =6\int \frac{1-t^3}{1+t^2}\cdot t^5 DT

\]

(Note: It has been converted into a rational function integral, when the molecular score is higher, and the polynomial division is used to calculate the true fraction.)

Because

\[

\frac{t^5-t^8}{1+t^2}=-t^6 +t^4 +t^3-t^2-t +1 + \frac{t-1}{1+t^2}.

\]

)

So the result is

\[

\begin{aligned}

\int \frac{1-\sqrt{x-1}}{1+\sqrt[3]{x-1}} dx =

-\frac (x-1) ^{7/6} +\frac (x-1) ^{5/6} + \FRAC32 (x-1) ^{2/3}-2 (x-1) ^{1/2}

-3 (x-1) ^{1/3}

\\+ 6 (x-1) ^{1/6} + 3 \ln| 1+ (x-1) ^{1/3} | -6\arctan (x-1) ^{1/6}+c

\end{aligned}

\]

(8) to make $t =\sqrt{e^x+1}$, the

\[

\int \frac{dx}{\sqrt{e^x+1}}=\int \frac{2}{t^2-1}dt==\ln|t-1|+\ln|t+1|+c=\ln|\sqrt{e^x+1}-1|-\ln|\sqrt{e^x+1}+1| +c.

\]

(9) (Universal transform most convenient) make $u =\tan \frac x2$

\[

\int \frac{dx}{\cos x +\sin x} =\int \frac{2}{1+2u-u^2} du =\int \frac{2}{2-(u-1) ^2}du=\frac{1}{\sqrt 2} \int (\frac{1} {\sqrt 2-(u-1)}+\frac{1}{\sqrt (u-1)}) du

\]

So

\[

\int \frac{dx}{\cos x +\sin x}=\frac 1{\sqrt 2} \left (-\ln| 1+\sqrt 2-\tan \frac x 2 |+ \ln | \sqrt 2-1 +\tan \FRAC X2 | \right) +c.

\]

(10) slightly

Assignment 23 indefinite integral of several kinds of special functions