Editorial - Matlan - Latihan Integral Eksponen dan ln

Problem 1

(∫_^)(ℇ(4*x+3)*d(x))=1/4*ℇ(4*x+3)+C

Problem 2

(∫_^)((2*x+ℇ(4*x+3))*d(x))=x2+1/4*ℇ(4*x+3)+C

Problem 3

u=2*x2+1

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

d(u)=4*x*d(x)=2⋅2*x*d(x)

(∫_^)(1/2*ℇu*d(x))=1/2*ℇu+C=1/2*ℇ(2*x2+1)+C

Problem 4

u=x2+x+1

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

d(u)=d(x)⋅(2*x+1)

(∫_^)(3)*(2*x+1)*ℇ(x2+x+1)*d(x)

(∫_^)(3⋅ℇu*d(u))=3*ℇ(x2+x+1)+C

Problem 5

(∫_^)((ℇ√(,2*x+1))/√(,8*x+4)*d(x))

u=√(,2*x+1)

d(u)/d(x)=1/√(,2*x+1)

d(x)=d(u)⋅√(,2*x+1)

(∫_^)((ℇu)/√(,4*(2*x+1))*d(u)⋅√(,2*x+1))

(∫_^)((ℇu)/(2√(,2*x+1))*d(u)⋅√(,2*x+1))

(∫_^)((ℇu)/2*d(u))=1/2*ℇ√(,2*x+1)+C

Problem 6

(∫_^)(1/(3*x+2)*d(x))

u=3*x+2

d(u)/d(x)=3

d(x)=d(u)/3

(∫_^)(1/ud(u)/3)=1/3*ln|3*x+2|+C

Problem 7

(∫_^)((3*x+5)/(3*x+1)*d(x))

(∫_^)(((3*x+1)+4)/(3*x+1)*d(x))=(∫_^)((1+4/(3*x+1))*d(x))

u=3*x+1

d(u)/d(x)=3

d(x)=d(u)/3

(∫_^)(1+4/ud(u)/3)=x+4/3*ln|3*x+1|+C

Problem 8

(∫_^)((6*x+1)/(2*x+3)*d(x))

(∫_^)((6*x+9-8)/(2*x+3))*d(x)

(∫_^)((3*(2*x+3)-8)/(2*x+3)*d(x))=(∫_^)((3-8/(2*x+3))*d(x))

u=2*x+3

d(u)/d(x)=2

d(x)=d(u)/2

(∫_^)(((3-8/u)*d(u))/2)=3/2*u-4*ln|2*x+3|+C

3/2⋅(2*x+3)-4*ln|2*x+3|+C

3*x+9/2-4*ln|2*x+3|+C

9/2 hilang karena konstanta masuk ke C. Sebenarnya seharusnya integral sebelumnya dipisah.

3*x-4*ln|2*x+3|+C

Problem 9

ok. get lucky

u=x2+x

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

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

(∫_^)((2*(2*x+1))/ud(u)/(2*x+1))=(∫_^)(2/u*d(u))=2*ln|x2+x|+C

Problem 10

u=x2+4*x+1

d(u)/d(x)=2*x+4

d(x)=d(u)/(2*x+4)

(∫_^)((3*(2*x+4))/ud(u)/(2*x+4))=(∫_^)(3/u*d(u))=3*ln|x2+4*x+1|+C

Problem 11

u=1-ℇx

d(u)/d(x)=-ℇx

d(x)=d(u)/(-ℇx)

(∫_^)(u3*ℇx⋅d(u)/(-ℇx))=(∫_^)(-u3*d(u))=-1/4*u4+C=-1/4*(1-ℇx)4+C

Problem 12

Direct u substitution won't work here.

(∫_^)(1/(ℇx+1)*d(x))=(∫_^)(((ℇx+1)-ℇx)/(ℇx+1))*d(x)=(∫_^)((1-(ℇx)/(ℇx+1))*d(x))

u=ℇx+1

d(u)/d(x)=ℇx

d(x)=d(u)/(ℇx)

(∫_^)(1*d(x))-(∫_^)((ℇx)/ud(u)/(ℇx))=x-ln|ℇx+1|+C

Problem 13

(ƒ^′)(x)=1+(x+2)/(x+4)=1+((x+4)-2)/(x+4)=1+1-2/(x+4)=2-2/(x+4)

ƒ(x)=(∫_^)((2-2/(x+4))*d(x))=2*x-2*ln|x+4|+C

Problem 14

(ƒ^′)(x)=(2*x+n)/(x2+4*x+3)

ƒ(x)=(∫_^)((2*x+n)/(x2+4*x+3)*d(x))

ok.

ƒ(x)=2*ln|x+1|+C=(∫_^)(2/(x+1)*d(x))

x2+4*x+3=(x+3)*(x+1)

(2*x+n)/((x+1)*(x+3))=2/(x+1)

(2*x+n)/(x+3)=2

2*x+n=2*x+6

∴n=6