# Integrals question

#### tmt

##### Active member

I am working on this question:

∫ [(3+ lnx)^2 (2-ln x)] / (4x) dx

18/4 [ 3 (ln x^2/2) + 4 (ln x^3/3) - ln x^4/4 ] + C

But the answer from the solutions is (5/12) (3+ ln x)^3 - (1/16)(3+lnx)^4 + C

Where did I mess up?

∫ [(3+ lnx)^2 (2-ln x)] / (4x) dx

let u =ln x

du/dx = 1/x

dx= du * x

substitute u into original equation,

∫ [(3+ u)^2 (2- u)] * (1/4x) * x du

x's cancel out

∫ [(3+ u)^2 (2- u)] * (1/4) du

remove constant

(1/4) ∫ [(3+ u)^2 (2- u)] du
expand

1/4 ∫ (9 + 6u + u^2) (2-u) du

1/4 ∫ 18 - 9u + 12u - 6u^2 + 2u^2 -u^3 du

simplify

1/4 ∫ 18 + 3u + 4u^2 - u^3 du

18/4 ∫ 3u + 4u^2 - u^3 du

integrate

18/4 [ 3 ∫ u du + 4 ∫ u^2 du - ∫ u^3 du ]

= 18/4 [ 3 (u^2/2) + 4 (u^3/3) - u^4/4 ] + c

replace u=lnx

= 18/4 [ 3 (ln x^2/2) + 4 (ln x^3/3) - ln x^4/4 ] + c

#### M R

##### Active member

I am working on this question:

∫ [(3+ lnx)^2 (2-ln x)] / (4x) dx

18/4 [ 3 (ln x^2/2) + 4 (ln x^3/3) - ln x^4/4 ] + C

But the answer from the solutions is (5/12) (3+ ln x)^3 - (1/16)(3+lnx)^4 + C

Where did I mess up?

∫ [(3+ lnx)^2 (2-ln x)] / (4x) dx

let u =ln x

du/dx = 1/x

dx= du * x

substitute u into original equation,

∫ [(3+ u)^2 (2- u)] * (1/4x) * x du

x's cancel out

∫ [(3+ u)^2 (2- u)] * (1/4) du

remove constant

(1/4) ∫ [(3+ u)^2 (2- u)] du
expand

1/4 ∫ (9 + 6u + u^2) (2-u) du

1/4 ∫ 18 - 9u + 12u - 6u^2 + 2u^2 -u^3 du

simplify

1/4 ∫ 18 + 3u + 4u^2 - u^3 du

18/4 ∫ 3u + 4u^2 - u^3 du

integrate

18/4 [ 3 ∫ u du + 4 ∫ u^2 du - ∫ u^3 du ]

= 18/4 [ 3 (u^2/2) + 4 (u^3/3) - u^4/4 ] + c

replace u=lnx

= 18/4 [ 3 (ln x^2/2) + 4 (ln x^3/3) - ln x^4/4 ] + c

It looks OK until you took the 18 outside the integral.

#### Prove It

##### Well-known member
MHB Math Helper

I am working on this question:

∫ [(3+ lnx)^2 (2-ln x)] / (4x) dx

18/4 [ 3 (ln x^2/2) + 4 (ln x^3/3) - ln x^4/4 ] + C

But the answer from the solutions is (5/12) (3+ ln x)^3 - (1/16)(3+lnx)^4 + C

Where did I mess up?

∫ [(3+ lnx)^2 (2-ln x)] / (4x) dx

let u =ln x

du/dx = 1/x

dx= du * x

substitute u into original equation,

∫ [(3+ u)^2 (2- u)] * (1/4x) * x du

x's cancel out

∫ [(3+ u)^2 (2- u)] * (1/4) du

remove constant

(1/4) ∫ [(3+ u)^2 (2- u)] du
expand

1/4 ∫ (9 + 6u + u^2) (2-u) du

1/4 ∫ 18 - 9u + 12u - 6u^2 + 2u^2 -u^3 du

simplify

1/4 ∫ 18 + 3u - 4u^2 - u^3 du

18/4 ∫ 3u - 4u^2 - u^3 du

integrate

18/4 [ 3 ∫ u du - 4 ∫ u^2 du - ∫ u^3 du ]

= 18/4 [ 3 (u^2/2) - 4 (u^3/3) - u^4/4 ] + c

replace u=lnx

= 18/4 { 3 [(ln x)^2/2] - 4 [(ln x)^3/3] - (ln x^4)/4 } + c

With the sign error fixed and the missing brackets inserted, this is correct.

Edit: Didn't notice the 18 being taken out of the integral. You can only do that with constant MULTIPLES.

#### Chris L T521

##### Well-known member
Staff member

I am working on this question:

∫ [(3+ lnx)^2 (2-ln x)] / (4x) dx

18/4 [ 3 (ln x^2/2) + 4 (ln x^3/3) - ln x^4/4 ] + C

But the answer from the solutions is (5/12) (3+ ln x)^3 - (1/16)(3+lnx)^4 + C

Where did I mess up?

∫ [(3+ lnx)^2 (2-ln x)] / (4x) dx

let u =ln x

du/dx = 1/x

dx= du * x

substitute u into original equation,

∫ [(3+ u)^2 (2- u)] * (1/4x) * x du

x's cancel out

∫ [(3+ u)^2 (2- u)] * (1/4) du

remove constant

(1/4) ∫ [(3+ u)^2 (2- u)] du
expand

1/4 ∫ (9 + 6u + u^2) (2-u) du

1/4 ∫ 18 - 9u + 12u - 6u^2 + 2u^2 -u^3 du

simplify

1/4 ∫ 18 + 3u + 4u^2 - u^3 du

18/4 ∫ 3u + 4u^2 - u^3 du

integrate

18/4 [ 3 ∫ u du + 4 ∫ u^2 du - ∫ u^3 du ]

= 18/4 [ 3 (u^2/2) + 4 (u^3/3) - u^4/4 ] + c

replace u=lnx

= 18/4 [ 3 (ln x^2/2) + 4 (ln x^3/3) - ln x^4/4 ] + c

I assume that they instead wanted you to make the substitution $u=3+\ln x\implies \ln x= u-3$ to minimize the amount of expanding you need to do in the integrand. It then follows that $x\,du =\,dx$ and thus your integral becomes

\begin{aligned}\frac{1}{4}\int u^2(5-u)\,du &= \frac{1}{4}\int 5u^2-u^3\,du\\ &= \frac{5}{12}u^3 - \frac{1}{16}u^4+C\end{aligned}

and thus $\displaystyle\int \frac{(3+\ln x)^2(2-\ln x)}{4x}\,dx = \frac{5}{12}(3+\ln x)^3 - \frac{1}{16}(3+\ln x)^4 + C$