Question

integration of x/(a+bx)​

Answer 1

Answer:

$\frac{x}{b} - \frac{a}{b^{2} } ln|a+bx| + C$

Step-by-step explanation:

let u = a + bx

⇒ $\frac{du}{dx}$ = b

∴ dx = $\frac{du}{b}$

∴ The integral becomes $\frac{1}{b} \int \frac{x}{u} \, du$

as u = a + bx

⇒ x = $\frac{u-a}{b}$

substituting x into the integral gives ⇒ $\frac{1}{b^{2} } \int \frac{u-a}{u} \, du$

⇒ $\frac{1}{b^{2} } \int ( 1 - \frac{a}{u} \ ) du$

⇒ $\frac{1}{b^{2} } ( u - a ln|u|) + C$

⇒$\frac{1}{b^{2} } ( a + bx - a ln|a +bx|) + C$

⇒ $\frac{a}{b^{2} } + \frac{x}{b} - \frac{a}{b^{2} } ln|a+bx| + C$

as $\frac{a}{b^{2} }$ is a constant, We can merge it with C

⇒ $\frac{x}{b} - \frac{a}{b^{2} } ln|a+bx| + C$

Hence above is the answer.

Hoping that I have not made any mistakes, You're welcome.

Hope you have found my answer useful. If my answer deserves a brainliest, do mark it.

GeniusH