Find the indefinite integral of 4x 2 1 under the root number...

Let x=sqrt(4*x 2+1)dx=

Then use the distribution integral (s is the integral number).

ssec(x)dtan(x)=ssec(x) 3dx, and then find ssec(x) 3dx

ssec(x)^3dx

ssec(x)dtan(x)

sec(x)*tan(x)-stan(x)dsec(x)sec(x)*tan(x)-stan(x)*(sec(x)^2-1)dx

sec(x)*tan(x)-ssec(x)^3+ssec(x)dxsec(x)*tan(x)-ssec(x)^3dx+ln|sec(x)+tan(x)|

See, there is ssec(x) 3 before and after, and you can just move the item in half.

It's too annoying on it, keep it simple.

Let x=then sqrt(4*x 2+1)dx=

Then it's like asking for it, it's simple.

Finally, let's talk about the answer.

asinh(2*x)/4 + x*(x^2 + 1/4)^(1/2)+c

Solution: f(x) under the root number (4x 2+1).

Reel 2x tan

f(x) under the root number (tan +1).

secθ|Because of secxdx

sec²x/secxdx

cosx/cos²xdx

1/cos²xdsinx

1/(1-sin²x)dsinx

1/(sinx+1)(sinx-1)dsinx-∫[1/(sinx-1)-1/(sinx+1)]/2dsinx-[∫1/(sinx-1)dsinx-∫1/(sinx+1)dsinx]/2

1/(sinx+1)d(sinx+1)-∫1/(sinx-1)d(sinx-1)]/2

ln|sinx+1|-ln|sinx-1|)/2+cln√|(sinx+1)/(sinx-1)|+cln√|(sinx+1)²/(sinx+1)(sinx-1)|+cln√|(sinx+1)²/(sin²x-1)|+cln√|-sinx+1)²/cos²x|+cln|(sinx+1)/cosx|+c

ln|tanx+1/cosx|+c

ln|secx+tanx|+c

Then substitute arctan2x.

secθ|＝2∫secθdθ

Let x=(tan t) 2

So it is integratively formed into sec t

d((tan t) 2)=sec 2 t 2 so the original formula = 1 2* sec 3 t dt=1 2* (1 cos 4 t)cos t dt

1 2* 1 (1-sin 2 t) 2 d(sint) and sint=u

Original = 1 2* 1 [(1-u) 2*(1+u) 2] du This is the integral of a rational function, and it should be.

When you're done, replace u with x.

It's a bit complicated.

Let x=2sint,t belong to the hermit hall eggplant in -pie 2 to pie 2,dx=2costdt,4-x 2=4[cost] 2=2cost

Just go down and make the jujubes. The flat stove of cost can be written in the form of [cos2t+1] 2.

Indefinite integral under the root number (1+4x 2):

Fractional Credits: UV).'=u'v+uv'

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

Both sides score: u'v dx=∫ (uv)' dx - uv' dx

i.e.: u'v dx = uv - uv'd, this is the partial integration formula.

It can also be abbreviated as: V du = UV - U DV

I suddenly remembered that this is a question in my high math midterm exam, and I can't solve it

Indefinite integral under the root number (1+4x 2):

Fractional Credits: UV).'=u'v+uv'

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

Points on both sides: u'v dx=∫ uv)' dx - uv' dx

i.e.: u'v dx = uv - uv'd, this is the Yunkuo division integral formula.

It can also be abbreviated as: V du = UV - U DV

Indefinite integral under the root number (1+4x 2):

Fractional Credits: UV).'=u'v+uv'

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

Points on both sides: u'v dx=∫ uv)' dx - uv' dx

i.e.: u'v dx = uv - uv'd, this is the Yunkuo division integral formula.

It can also be abbreviated as: V du = UV - U DV

1/√4-9x^2

1/2√[1-(3x/2)^2]dx

1 3) Answer: Lu 1 [1-(3x 2) 2] d(3x 2)(1 stupid bump 3)arcsin(3x 2)+c to talk about the answers are for reference only, if you have any questions, you can continue to ask!