Respuesta :
Answer:
[tex]\text{ln}(5x)(4x^2+10x)-2x^{2}-10x+C[/tex]
Step-by-step explanation:
We have been given an definite integral [tex]\int \left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx[/tex]. We are asked to find the integral using integration by parts.
We will use Integration by parts formula to solve our given problem.
[tex]\int\ vdv=uv-\int\ vdu[/tex]
Let [tex]u=\text{ln}(5x)[/tex] and [tex]v'=8x+10[/tex].
Now, we need to find du and v using these values as shown below:
[tex]\frac{du}{dx}=\frac{du}{dx}(\text{ln}(5x))[/tex]
Using chain rule, we will get:
[tex]\frac{du}{dx}=\frac{1}{5x}*5[/tex]
[tex]\frac{du}{dx}=\frac{1}{x}[/tex]
[tex]du=\frac{1}{x}dx[/tex]
[tex]v'=8x+10[/tex]
[tex]v=\frac{8x^{1+1}}{2}+10x[/tex]
[tex]v=\frac{8x^{2}}{2}+10x[/tex]
[tex]v=4x^{2}+10x[/tex]
Upon substituting these values in integration by parts formula, we will get:
[tex]\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-\int\ (4x^2+10x)\frac{1}{x}dx[/tex]
[tex]\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-\int\ \frac{4x^2}{x}+\frac{10x}{x}dx[/tex]
[tex]\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-\int 4x+10dx[/tex]
[tex]\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-(\frac{4x^{2}}{2}+10x)+C[/tex]
[tex]\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-2x^{2}-10x+C[/tex]
Therefore, our required integral would be [tex]\text{ln}(5x)(4x^2+10x)-2x^{2}-10x+C[/tex].
