x dx 1 2x dx 1 d(7+x²) 1
∫ = ∫ = ∫ = ln(7+x²)+C
7+x² 2 7+x² 2 7+x² 2
[1/2 *ln(7+x²)+C ]¹= 1/2*[ 2x /(7+x²)+0]= x /(7+x²)
x+18 (x-2)+20 1 2(x-2) dx
2) ∫dx=∫ dx= ∫ dx+20 ∫ =
x²-4x-12 (x-2)²-16 2 (x-2)²-16 (x-2)²-16
1 1 | x-2-4 | 1 5 | x-6 |
= *ln|(x-2)²-16|+20 * *ln || +C= *ln |x²-4x-12|+*ln || +C
2 2*8 | x-2+4 | 2 4 | x+2 |
3) ∫(3-x) cosx dx=[ u=3-x , du=-dx , dv=cosx dx , v=sinx ] =(3-x)sinx+∫ sinx dx=
=(3-x)sinx-cosx+C
[(3-x)sinx-cosx]¹= -sinx+(3-x)cosx+sinx +0=(3-x)cosx
149,6:(1,36+1,6у)=112,5-12,5
149,6:(1,36+1,6у)=100
1,36+1,6у=149,6:100
1,36+1,6у=1,496
1,6у=1,496-1,36
1,6у=0,136
у=0,136:1,6
у=0,085
Проверка:
12,5+149,6:(1,36+1,6*0,085)=112,5
12,5+149,6:(1,36+0,136)=112,5
12,5+149,6:1,496=112,5
12,5+100=112,5
112,5=112,5