Question

Exercise is mixed —some require integration by parts, while others can be integrated by using techniques discussed in the chapter on Integration.
∫(8x+10) ln (5x) dx.

Answer 1

Answer:

$\text{ln}(5x)(4x^2+10x)-2x^{2}-10x+C$

Step-by-step explanation:

We have been given an definite integral $\int \left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx$. We are asked to find the integral using integration by parts.

We will use Integration by parts formula to solve our given problem.

$\int\ vdv=uv-\int\ vdu$

Let $u=\text{ln}(5x)$ and $v'=8x+10$.

Now, we need to find du and v using these values as shown below:

$\frac{du}{dx}=\frac{du}{dx}(\text{ln}(5x))$

Using chain rule, we will get:

$\frac{du}{dx}=\frac{1}{5x}*5$

$\frac{du}{dx}=\frac{1}{x}$

$du=\frac{1}{x}dx$

$v'=8x+10$

$v=\frac{8x^{1+1}}{2}+10x$

$v=\frac{8x^{2}}{2}+10x$

$v=4x^{2}+10x$

Upon substituting these values in integration by parts formula, we will get:

$\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-\int\ (4x^2+10x)\frac{1}{x}dx$

$\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-\int\ \frac{4x^2}{x}+\frac{10x}{x}dx$

$\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-\int 4x+10dx$

$\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-(\frac{4x^{2}}{2}+10x)+C$

$\int\left(8x+10\right)\:\text{ln}\:\left(5x\right)\:dx=\text{ln}(5x)(4x^2+10x)-2x^{2}-10x+C$

Therefore, our required integral would be $\text{ln}(5x)(4x^2+10x)-2x^{2}-10x+C$.