Question

I=integral dx upon x+2√x+3​

Answer 1

Step-by-step explanation:

let, x+3 = z² ► dx = 2z dz

so,

$\int\limits {\frac{2z}{z^2-3+ 2z} } \, dz$

here d({z²-3+ 2z) = (2z+2) dz\

$\int\limits {\frac{2z + 2 - 2}{z^2-3+ 2z} } \, dz$

= $\int\limits {\frac{d(z^2-3+ 2z)}{z^2-3+ 2z} } - 2 \int\limits {\frac{1}{z^2-3+ 2z} } \, dz$

=  $ln(z^2 - 3 + 2z)$ $- 2 \int\limits {\frac{1}{z^2 +3z - z-3} } \, dz$

=  $ln(z^2 - 3 + 2z)$ $- 2 \int\limits {\frac{1}{(z+3)(z-1)} } \, dz$

=  $ln(z^2 - 3 + 2z) - 2\int\limits {(\frac{1}{-4(z+3)} +\frac{1}{4(z-1)}} )\, dz$

=  $ln(z^2 - 3 + 2z) + \int\limits {(\frac{1}{2(z+3)} -\frac{1}{2(z-1)}} )\, dz$

=  $ln(z^2 - 3 + 2z) + \frac{1}{2} ln(z+3) - \frac{1}{2} ln(z-1)$

= $ln(|x + 2\sqrt{x+3}|) + \frac{1}{2} ln(|\sqrt{x+3}+3|) - \frac{1}{2} ln(|\sqrt{x+3}-1|)$