Editorial - Matlan - Latihan Integral 2

Problem 1

(∫_^)(21*(7*x-9)11*d(x))

u=7*x-9

d(u)/d(x)=7

d(x)=d(u)/7

(∫_^)(21*u11⋅d(u)/7)=1/4*u12+C=1/4*(7*x-9)12+C

Problem 2

(∫_^)(25*x*(5*x-3)4*d(x))

u=5*x-3

x=(u+3)/5

d(u)/d(x)=5

d(x)=d(u)/5

(∫_^)(25⋅(u+3)/5⋅u4⋅d(u)/5)=(∫_^)((u5+3*u4)*d(u))

1/6*u6+3/5*u5=1/6*(5*x-3)6+3/5*(5*x-3)5+C

Problem 3

(∫_^)(x√(,16-x2)*d(x))

u=16-x2

x2=16-u

d(u)/d(x)=-2*x

x*d(x)=d(u)/(-2)

(∫_^)(√(,u)⋅d(u)/(-2))=-1/2⋅1/3/2*u(3/2)+C=(-u3/2)/3+C=-1/3*(16-x2)(3/2)+C

Problem 4

(∫_^)(8*x√(3,2*x2+5)*d(x))

u=2*x2+5

d(u)/d(x)=4*x

x*d(x)=d(u)/4

(∫_^)(8⋅u(1/3)⋅d(u)/4)=2⋅3/4⋅u(4/3)+C=3/2*(2*x2+5)(4/3)+C

Problem 5

(∫_^)((6*x2)/√(,x3-27)*d(x))

u=x3-27

d(u)/d(x)=3*x2

x2*d(x)=d(u)/3

(∫_^)(6/(u(1/2))⋅d(u)/3)=2⋅2⋅u(1/2)+C=4⋅(x3-27)(1/2)+C

Problem 6

(∫_^)(x/(x2+1)*d(x))

u=x2+1

d(u)/d(x)=2*x

x*d(x)=d(u)/2

(∫_^)(u(-1)⋅1/2⋅d(u))

1/2*(∫_^)(1/u*d(u))

1/2*ln(u)+C=1/2*ln|x2+1|+C

Problem 7

(∫_^)((8*x-4)/√(3,(x-3)*(x+2))*d(x))

(x-3)*(x+2)=x2-x-6

u=x2-x-6

d(u)/d(x)=2*x-1

d(u)=(2*x-1)*d(x)

(∫_^)((4*(2*x-1))/√(3,x2-x-6)*d(x))=(∫_^)(4*u(-1/3)*d(u))=6*u2/3+C=6*(x2-x-6)(2/3)+C

Problem 8

(∫_^)(((9*x)*(x-10))/√(4,x3-15*x2-16)*d(x))

u=x3-15*x2-16

d(u)/d(x)=3*x2-30*x=3*x*(x-10)

d(u)=3*x*(x-10)*d(x)

(∫_^)(3⋅u(-1/4)*d(u))

3⋅4/3⋅u(3/4)+C=4*(x3-15*x2-16)(3/4)+C

Problem 9

(∫_^)(9/2*x2√(,x3+1)*d(x))

u=x3+1

d(u)/d(x)=3*x2

d(u)=3*x2*d(x)

(∫_^)(3/2*u(1/2)*d(u))

3/2⋅2/3⋅u(3/2)+C=(x3+1)(3/2)+C

Problem 10

ƒ(x)=(∫_^)((6*x+3)*(x2+x)2*d(x))

ƒ(1)=8

d(u)/d(x)=2*x+1

ƒ(x)=(∫_^)(3*u2)=(x2+x)3+C

ƒ(1)=8+C=8,∴C=0

ƒ(x)=(x2+x)3