Although this integral might not seem like this, you need to use both the u-substitution method and integration by parts to evalutate this integral. First, I'll use the u-substitution method to make the integral simpler.
Let A = 5x - 4 , dA = 5dx,
∫ ln(5x - 4) dx
= ∫ ln(A) dx
= ∫ ln(A) * 1/5 * 5dx
= ∫ ln(A) * 1/5 * dA
= 1/5 ∫ ln(A) dA
Now, let's evaluate the antiderivative of ln(A) using integration by parts (∫u dv = uv - ∫v du).
Let u = ln(A), du = 1/A dA,
Let dv = dA , v = A,
1/5 ∫ ln(A) dA
= 1/5 [ ln(A) * A - ∫A * 1/A dA ]
= 1/5 [ A*ln(A) - ∫1 dA ]
= 1/5 [ A*ln(A) - A + C]
Afterwards, by substituting 5x - 4 back to A, we can get the final answer, which would be:
Hi,
Thank you for your question.
Although this integral might not seem like this, you need to use both the u-substitution method and integration by parts to evalutate this integral. First, I'll use the u-substitution method to make the integral simpler.
Let A = 5x - 4 , dA = 5dx,
∫ ln(5x - 4) dx
= ∫ ln(A) dx
= ∫ ln(A) * 1/5 * 5dx
= ∫ ln(A) * 1/5 * dA
= 1/5 ∫ ln(A) dA
Now, let's evaluate the antiderivative of ln(A) using integration by parts (∫u dv = uv - ∫v du).
Let u = ln(A), du = 1/A dA,
Let dv = dA , v = A,
1/5 ∫ ln(A) dA
= 1/5 [ ln(A) * A - ∫A * 1/A dA ]
= 1/5 [ A*ln(A) - ∫1 dA ]
= 1/5 [ A*ln(A) - A + C]
Afterwards, by substituting 5x - 4 back to A, we can get the final answer, which would be:
1/5 [ (5x - 4) * ln(5x - 4) - (5x - 4) + C]
= ( (5x - 4) / 5) * ln(5x - 4) - (5x - 4) / 5 + C
Does this answer your question?