How do you integrate $\int x^{3} {ln x}^{4} d x$ $int x^3 ln x^4 dx$ using integration by parts?

Question

How do you integrate $\int x^{3} {ln x}^{4} d x$ $int x^3 ln x^4 dx$ using integration by parts?

2 Answers

Impact of this question

6917 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-01T04:21:02

$\int x^{3} ln (x^{4}) d x = x^{4} ln (x) - \frac{1}{4} x^{4} + C$ $intx^3ln(x^4)dx=x^4ln(x)-1/4x^4+C$

Explanation:

First, let's rewrite $ln (x^{4})$ $ln(x^4)$ as $4 ln (x),$ $4ln(x),$ as $ln (x^{a}) = a ln (x)$ $ln(x^a)=aln(x)$

$\int x^{3} ln (x^{4}) d x = \int 4 x^{3} ln (x) d x = 4 \int x^{3} ln (x) d x$ $intx^3ln(x^4)dx=int4x^3ln(x)dx=4intx^3ln(x)dx$ (We can factor out $4,$ $4,$ it's just a constant.)

Make the following choices for $u$ $u$ and $d v :$ $dv:$

$u = ln (x)$ $u=ln(x)$

$d v = x^{3} d x$ $dv=x^3dx$

Thus, $d u$ $du$ and $v$ $v$ become:

$d u = \frac{1}{x} d x$ $du=1/xdx$

$d v = \int x^{3} d x = \frac{1}{4} x^{4}$ $dv=intx^3dx=1/4x^4$ (We're not going to include the constant just yet. It can wait until the end.)

Now it becomes apparent why we made the choices we did for $u$ $u$ and $v .$ $v.$ If we chose $d v = ln (x),$ $dv=ln(x),$ we would have to integrate $ln (x)$ $ln(x)$ for $v,$ $v,$ and would run into our original problem.

Plug in relevant values:

$u v - \int v d u = \frac{1}{4} x^{4} ln (x) - \int \frac{1}{4} \frac{x^{4}}{x} d x$ $uv-intvdu=1/4x^4ln(x)-int1/4x^4/xdx$

We can simplify the integral here:

$1/4x^4ln(x)-int1/4x^4/xdx=1/4x^4ln(x)-1/4intx^(cancel(4)3)/cancelx$

Thus, we have

$1/4x^4ln(x)-1/4intx^3dx=1/4x^4ln(x)-(1/4)(1/4)x^4+C=1/4x^4ln(x)-1/16x^4+C$

Let's not forget the $4$ we originally factored out...

$4(1/4x^4ln(x)-1/16x^4+C)=x^4ln(x)-1/4x^4+C$

$C$ remains unchanged; a constant multiplied by a constant remains a constant.

So,

$intx^3ln(x^4)dx=x^4ln(x)-1/4x^4+C$

Answer 2 · 2018-03-01T04:51:53

$int\ x^3ln(x^4)\ "d"x=4x^4ln(x)-x^4+C$

Explanation:

Here's another approach:

Consider substituting first $w=x^4$ and $("d"w)/("d"x)=4x^3$ into the integral $int\ x^3ln(x^4)\ "d"x$ .

This gives $1/4int\ ln(w)\ "d"w$ .

Now use integration by parts $int\ u\ "d"v=uv-int\ v\ "d"u$ . Use $u=ln(w)$ and $"d"v="d"w$ . Then find that $"d"u=("d"w)/w$ and $v=w$ .

Now, we have $1/4int\ ln(w)\ "d"w=1/4(wln(w)-int\ "d"w)$
or simply $1/4wln(w)-1/4w+C$ .

Substitute back $w=x^4$ .

Our final answer is thus $1/4x^4ln(x^4)-1/4x^4+C=x^4ln(x)-1/4x^4+C$ .

Science

Math

Humanities

... and beyond

How do you integrate $\int x^{3} {ln x}^{4} d x$ $int x^3 ln x^4 dx$ using integration by parts?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate int x^3 ln x^4 dx ∫x3lnx4dxint x^3 ln x^4 dx using integration by parts?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you integrate $\int x^{3} {ln x}^{4} d x$ $int x^3 ln x^4 dx$ using integration by parts?