Anonymous

Let 8+3x=u, then 3dx=du Sdx/(8+3u)= Sdu/[3u]= ln(u)/3+C= ln(8+3x)/3+C where 8+3x>0 or x>-8/3 for the real solution.

Anonymous

ln|8 + 3x|/3 + c

Anonymous

Let u = 8 + 3x du/dx = 3 3dx = du dx = du/3 ∫ (1/3u du (1/3) ∫ (1/u) du (1/3) ∫ ln u (1/3) ln |8 + 3x| ln |8 + 3x|/3 + C

Anonymous

Let u=8+3x du = 3 dx dx = (1/3) du ∫(dx/(8+3x)) = (1/3) ∫ du/u = (1/3) ln |u| replace u by 8+3x = (1/3) ln |8+3x| + C

Anonymous

. u = 8 + 3x —————> du/dx = 3 ————> dx = ⅓ du ∫(dx/(8+3x)) = ∫1/(8 + 3x) dx = ∫ 1/u * ⅓ du = ⅓ ∫1/u du = ⅓ ln|u| + C = ⅓ ln|8 + 3x| + C ————————— or you can express the constant as a natural log ⅓ ln|8 + 3x| + ⅓ln|c| = ⅓ ln|c (8 + 3x) | ————————

Anonymous

Put J = Int'l[1/(8+3x)]dx. Substitute u for 8+3x. Then du/dx = 3, ie., dx = (1/3)du & J = (1/3)Int'l[1/u]du = (1/3)ln(u) +c = (1/3)ln(3x+8) +c.

Anonymous

u = 8 + 3 x leads to (1/3) ln ( 3x + 8 ) + C

Anonymous

u = 8 + 3x du = 3 * dx (1/3) * du = dx dx / (8 + 3x) => (1/3) * du / u Integrate (1/3) * ln|u| + C => (1/3) * ln|8 + 3x| + C

Anonymous

t=8+3x dt=3*dx dx=(1/3)dt hence =(1/3)*∫1/t dt=(1/3)*log|t| +c= =(1/3)log|8+3x| +c

Anonymous

u = 8 + 3x du = 3dx dx = 1/3 * du ∫ 1/3 * du / u can you take it from there?