Use the partial integration method to find sinx e ax dx

(x2^x)/in2-2^x/(ln^2x)

The partial integral method is as follows:

x2^xdx

1/ln2)∫xd2^x

x2^x)/ln2-(1/ln2)∫2^xdx

x2^x)/in2-2^x/(ln^2x)

Formula for indefinite integrals.

1. A dx = ax + c, a and c are constants.

2. x a dx = [x (a + 1)] (a + 1) +c, where a is a constant and a ≠ 1

3、∫ 1/x dx = ln|x| +c

4. A x DX = (1 LNA)a x + C, where A > 0 and A ≠ 1

5、∫ e^x dx = e^x + c

6、∫ cosx dx = sinx + c

7、∫ sinx dx = - cosx + c

8、∫ cotx dx = ln|sinx| +c = - ln|cscx| +c

The final result of finding the indefinite integral x2 xdx using the partial integration method is (x2 x) in2-2 x (ln 2x).

x2^xdx

1/ln2)∫xd2^x

x2^x)/ln2-(1/ln2)∫2^xdx=(x2^x)/in2-2^x/(ln^2x)。

So the final result of finding the indefinite integral x2 xdx using the partial integration method is (x2 x) in2-2 x (ln 2x).

The partial integral method is as follows:

x2^xdx

1/ln2)∫xd2^x

x2 x) ln2-(1 ln2) 2 xdx=(x2 x) in2-2 x (ln 2x) indefinite integral.

1, a dx = ax + c, a and c are both constants2, x a dx = [x (a + 1)] (a + 1) +c, where a is a constant and a ≠ 1

3、∫ 1/x dx = ln|x|+c4, a x dx = (1 lna)a x + c, where a > 0 and a ≠ 1

5、∫ e^x dx = e^x + c

6、∫ cosx dx = sinx + c7、∫ sinx dx = - cosx + c8、∫ cotx dx = ln|sinx| +c = - ln|cscx| +c

x e xdx=x e x-2xe x+2e x+ is constant.

x²e^xdx

x²d(e^x)

x²e^x-∫e^xd(x²)

x²e^x-∫2xd(e^x)

x²e^x-2xe^x+∫2d(e^x)=x²e^x-2xe^x+2e^x+c

Extended Information: Segment Credits:

uv)'=u'v+uv'

Got:u'v=(uv)'-uv'

Points on both sides get:

u'vdx=∫

uv)'dx

uv'dx stands for: u'vdx

uvuv'd, this is the partial integration formula.

It can also be abbreviated as: VDU

uvudv common integration formula:

1）∫0dx=c

2）∫x^udx=(x^(u+1))/u+1)+c3）∫1/xdx=ln|x|+c

4）∫a^xdx=(a^x)/lna+c

5）∫e^xdx=e^x+c

6）∫sinxdx=-cosx+c

7）∫cosxdx=sinx+c

8）∫1/(cosx)^2dx=tanx+c9）∫1/(sinx)^2dx=-cotx+c10）∫1/√（1-x^2)

dx=arcsinx+c

Summary. Hello, please send out the complete question Ha Partial Integral Method to Find Indefinite Integral Calendar: (1) x e Search x dx (2) e (2x) cosx dx Trouble cavity letter writing paper.

Hello, please send out the full question

Use the partial integral method to find the indefinite integrals of these two problems.

It's not easy to count <>

There is one more question.

There's also a question.

The second one I didn't expect to find out directly, the answer can be obtained twice by partial integration, which is more special.

The original formula is permeable = - xde (-x).

xe (-x) + shouting missing e (-x)dx

xe^(-x)-e^(-x) (1,0)(-1/e-1/e)-(0-1)

1-2 pure oak

Original =- xde (-x).

xe^(-x)+∫e^(-x)dx

xe (-x)-e (-x) (1,0)(-1 Shenhuyan e-1 e)-(0-1).

1-2 You Yu E,5, why is it not used (b,a)udv=uv|(b,a)- b,a)vdu is in the formula,

False, correct as follows:

m=∫e^(-2x)sin(x/2)dx

1/2)∫sin(x/2)d[e^(-2x)]

1/2)sin(x/2)e^(-2x)-(1/2)∫e^(-2x)d[sin(x/2)]

1/2)sin(x/2)e^(-2x)+(1/4)∫e^(-2x)cos(x/2)dx

1/2)sin(x/2)e^(-2x)+(1/8)∫cos(x/2)d[e^(-2x)]

1/2)sin(x/2)e^(-2x)+(1/8)cos(x/2)e^(-2x)+(1/8)∫e^(-2x)dcos(x/2)

1/2)sin(x/2)e^(-2x)+(1/8)cos(x/2)e^(-2x)+(1/16)∫e^(-2x)sin(x/2)dx

1/2)sin(x/2)e^(-2x)+(1/8)cos(x/2)e^(-2x)+(1/16)m

Thus m=(16 17)*[1 2)sin(x 2)e (-2x)+(1 8)cos(x 2)e (-2x)]+c

2/17)e^(-2x)[ 4sin(x/2) +cos(x/2)] c