Question

Integral x^2/1+5x dx

Answer 1

step by step solution

Answer 2

$\int {\dfrac{x^2}{1+5x} } \, dx=\dfrac{1}{125} (\dfrac{t^2}{2}-2t+\log [t]+c )$

Step-by-step explanation:

Let$I=\int {\dfrac{x^2}{1+5x} } \, dx$

To find, $\int {\dfrac{x^2}{1+5x} } \, dx=?$

∴ $I=\int {\dfrac{x^2}{1+5x} } \, dx$

Let 1 + 5x = t

⇒ 5dx = dt ⇒ dx = $\dfrac{dt}{5}$

∴ $x=\dfrac{t-1}{5}$

$I=\dfrac{1}{5} \int {\dfrac{(\dfrac{t-1}{5})^2}{t} } \, {dt}$

$I=\dfrac{1}{5} \int {\dfrac{(\dfrac{t^2-2t+1}{25}}{t} } \, {dt}$

$I=\dfrac{1}{125} \int {(t-2+\dfrac{1}{t}}) \, {dt}$

$I=\dfrac{1}{125} (\dfrac{t^2}{2}-2t+\log [t]+c )$

Where, c is called integration constant

Hence, $\int {\dfrac{x^2}{1+5x} } \, dx=\dfrac{1}{125} (\dfrac{t^2}{2}-2t+\log [t]+c )$.