dx x 2 4x indefinite integrals

Original = dx (x 2-4x+4-4).

d(x-2) [(x-2) 2-4] (becomes derivative(x-2), the formula remains unchanged).

Let x-2=2sect, then x=2sect+2, so dx=2sect*tantdt

So the original = dt [4((sect) 2-1] *2sect*tant

dt/(2tant) *sect

1/2*∫dt/sint

1 2* dt*sint (sint) 2 (up and down multiplied by one sint).

1 2* dcost [1-(cost) 2] (put the sint inside d).

1 4*[ dcost (1-cost) +dcost (1+cost)] splits the denominator, and at the same time raises 1 2 out).

1/4*[-ln(1-t)+ln(1+t)+c]

ln[(1+t)/(1-t)]+c

ln[(1+arccos(2/(x-2)))/(1-arccos(2/(x-2)))c

Convert the previous 2sect=x-2 to the form of x representing t, and you are done. )

Method 1: Use the formula dx (a +b x) = (1 ab)arctan(bx a) +c

dx/(x² +4) = (1/2)arctan(x/2) +c

Method 2: Trigonometric function commutation: let x = 2tanz, dx = 2sec z dz

dx/(x² +4)

2sec²z dz)/(4tan²z + 4)

2sec²z/[4(tan²z + 1)] dz

1/2)∫ sec²z/sec²z dz

z/2 + c

1 2) arctan(x2) +c, because tanz = x2

We can divide the integrand into two parts:

j = x^2 + 4x)^(1)]dx = 1/x^2)dx - 4/x(x+4))dx

The first integral can be solved directly with the indefinite integral formula of the power function:

1/x^2)dx = 1/x + c1

where c1 is a constant.

The second integral can be split into two parts:

4 x(x+4))dx = 1 x)dx - 1 (x+4))dx The first part of dx can be solved using the indefinite dross ridge fraction formula of the logarithmic function:

1/x)dx = ln|x| +c2

where c2 is a constant.

The second part can be solved using the commutation method.

Let u = x + 4, then du dx = 1 and dx = du. Substitute the original formula to get:

1/(x+4))dx = 1/u)du = ln|u|+c3 where c3 is a constant.

Replace you back with x to get:

1/(x+4))dx = ln|x+4|+c3 So, adding these two parts, we get:

j = 1/x + ln|x| +ln|x+4|+c, where c is a constant.

Therefore, the indefinite integral of j is:

x^2 + 4x)^(1)]dx = 1/x + ln|x| +ln|x+4| +c

First of all, we can decompose the function of the accumulation beam finger to get:

x²-4x)/dx = x²+4x)/dx = x²)/dx + 4x)/dx

Now we can solve for each indefinite integral:

x²)/dx = x³/3 + c1

where c1 denotes an arbitrary constant.

4x) dx = 4 (x) dx = 4x + c2, where c2 denotes the constant of any heart.

Add the two integral results together to get the final indefinite integral:

x²-4x)/dx = x²)/dx + 4x)/dx = x³/3 + 4x + c

where c = c1 + c2 represents any constant.

Distribute the accumulation of Xun Xun first, Liquid Changkai ln(x 2+4)dx=xln(x 2+4)- xdln(x 2+4)=xln(x 2+4)- 2x 2 (x 2+4)]dx=xln(x 2+4)-2 [x 2 (x 2+4)]dx=xln(x 2+4)-2 [(x 2+4-4) (x 2+4)]dx=xln(x 2+4)-2 [1-4 (x 2+4)]dx=xln(x 2+4)-2x+2 [4]. (x 2...

The root number is squared and rented, x=x-2+2

DX also made up the corresponding square number of drawbacks.

Then split into two parts.

One is the integral of a power function and the other is the integral of a circle.

That's the idea