Question

integrate the following​

Answer 1

Answer:

$\int \dfrac{6}{2x+5}dx\\\\\implies 6.\int \dfrac{1}{2x+5}dx\\\\\textbf{Let 2x+5 be u}\\\\\implies 6.\int \dfrac{1}{2u}du\:\:[\dfrac{d}{dx}[2x+5]=2]\\\\\implies 6 \times \dfrac{1}{2} \int \dfrac{1}{u}du\\\\\implies 3 log |u|\\\\\implies \boxed{3log|2x+5|+C}$

C is integral constant .

Explanation:

Substitute 2 x + 5 with another variable u to make it faster .

The derivative of 2 x + 5 is 2 and hence we take $\int \dfrac{1}{2u}du=\int \dfrac{1}{2x+5}dx$ where 2 x + 5 = u .

$2x+5=u\\\\\implies \dfrac{dx}{d}[2x+5]=\dfrac{du}{d}\\\\\implies 2\dfrac{dx}{d}=\dfrac{du}{d}\\\\\implies dx=\dfrac{du}{2}$

Answer 2

SOLUTION

We have,

$\displaystyle{ \sf{l = \int \: \dfrac{6}{2x + 5} \: dx }}$

Rewriting the equation as :

$\longrightarrow \: \displaystyle{ \sf \: l = 6 \int \: \dfrac{1}{2x + 5}dx }$

Assume,

$\sf u = 2x + 5$

Taking derivative both sides,

$\leadsto \: \sf \: du = 2dx \\ \\ \leadsto \: \sf \: dx = \dfrac{du}{2}$

Now,

$\displaystyle{ \longrightarrow \: \sf \: l = 6 \int \: \dfrac{1}{u} . \dfrac{du}{2} } \\ \\ \longrightarrow \: \displaystyle \: \sf \: l = \dfrac{6}{2} log(u) + c \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \: l = 3log(2x + 5) + c}}$

Using the Power Rule of Logarithms,

$\longrightarrow \displaystyle \sf I = log((2x+5)^3) + c$