Question

help please. How can I integrate this one?​

Answer 1

$\implies\displaystyle \int {(2x + 3)}^{2} dx$

We know that :-

• (a+b)²= a²+b²+2ab

$\implies\displaystyle \int ( 4 {x}^{2} + 9 + 12x \: )dx$

$\implies\displaystyle \int 4 {x}^{2} dx + \displaystyle \int9 \: dx + \displaystyle \int12x \: dx$

$\implies \dfrac{4 {x}^{3} }{3} + 9x + \dfrac{12 {x}^{2} }{2} + c$

where c is constant

$\implies \dfrac{4 {x}^{3} }{3} + 9x + \dfrac{6{x}^{2} }{1} + c$

$\implies \dfrac{4 {x}^{3} }{3} + 9x + 6{x}^{2} + c$

Answer :

$\dfrac{4 {x}^{3} }{3} + 9x + 6{x}^{2} + c$

____________________

$\large\bf\red{Learn\:More}$

∫ 1 dx = x + C

∫ sin x dx = – cos x + C

∫ cos x dx = sin x + C

∫ sec2 dx = tan x + C

∫ csc2 dx = -cot x + C

∫ sec x (tan x) dx = sec x + C

∫ csc x ( cot x) dx = – csc x + C

∫ (1/x) dx = ln |x| + C

∫ ex dx = ex+ C

∫ ax dx = (ax/ln a) + C

______________________

Answer 2

QuestioN :

$\tt \bullet\int( 2x + 3 )^2 \,dx$

AnsWer :

4x²/3 + 9x + 6x² + c.

SolutioN :

Integration w.r.t.x

$\tt \mapsto I = \int( 2x + 3 )^2 \,dx$

$\tt \mapsto I = \int(4 {x}^{2} + 9 + 12x) \,dx$

★ By Sum rule of integration separate the terms.

$\tt \mapsto I = \int4 {x}^{2}\,dx + \int 9\,dx + \int 12x\,dx$

$\tt \mapsto I = 4\int {x}^{2}\,dx + 9\int \,dx + 12\int x\,dx$

$\tt \mapsto I = 4\frac{ {x}^{3}}{3} + 9x + \cancel{12}\frac{ {x}^{2} }{\cancel2} + c.$

$\tt \mapsto I = \frac{ {4x}^{3}}{3} + 9x + 6{x}^{2} + c.$

Therefore, the required value be 4x²/3 + 9x + 6x² + c.

MorE InformatioN.

$\tt \mapsto 1\int\,dx = x + c.$

$\tt \mapsto \int {x}^{n} \,dx = \frac{{x}^{n + 1}}{n + 1} + c.$

$\tt \mapsto \int\,Cosx\,dx = Sinx + c.$

$\tt \mapsto \int\,Sinx\,dx = -Cosx + c.$

$\tt \mapsto \int\,Sec^2x\,dx = tanx + c.$