Question

integration of [√x(3x² + 2x  + 3)].dx

Answer 1

$\bf{I=\int{[\sqrt{x}(3x^2+2x+3)]}\,dx}\\\\=\bf{\int{[3x^2.\sqrt{x}+2x.\sqrt{x}+3\sqrt{x}]}\,dx}\\\\=\bf{\int{[3x^{5/2}+2x^{3/2}+3x^{1/2}]}\,dx}\\\\=\bf{3\int{x^{5/2}}\,dx+2\int{x^{3/2}}\,dx+3\int{x^{1/2}}\,dx}\\\\=\bf{3.\frac{x^{5/2+1}}{5/2+1}+2\frac{x^{3/2+1}}{3/2+1}+3\frac{x^{1/2+1}}{1/2+1}}\\\\=\bf{\frac{6}{7}x^{7/2}+\frac{4}{5}x^{5/2}+2x^{3/2}+C}$

hence, I = $\bf{\frac{6}{7}x^{7/2}+\frac{4}{5}x^{5/2}+2x^{3/2}+C}$

Answer 2

HELLO DEAR,

given function is I = ∫√x(3x² + 2x + 3).dx

⇒∫(3x²*√x + 2x*√x + 3√x).dx

⇒∫(3x^{5/2} + 2x^{3/2} + 3x^{1/2}).dx

⇒3∫x^{5/2}.dx + 2∫x^{3/2}.dx + 3∫x^{1/2}.dx

⇒$\bold{3\frac{x^{7/2}}{7/2} + 2\frac{x^{5/2}}{5/2} + 3\frac{x^{3/2}}{3/2} + c}$

⇒$\bold{\frac{6x^{7/2}}{7} + \frac{4x^{5/2}}{5} + \frac{6x^{3/2}}{3} + c}$

hence, $\bold{I = \frac{6x^{7/2}}{7} + \frac{4x^{5/2}}{5} + {2x^{3/2}} + c}$

I HOPE ITS HELP YOU DEAR,
THANKS