Question

solve the problem with explanation ✋✋​

Answer 1

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.

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

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

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

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

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

Answer 2

${\fbox{\fbox{\huge\sf\purple{QuEsTiOn}}}}$

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

AnsWer :

4x²/3 + 9x + 6x² + c.

${\fbox{\fbox{\huge\sf\purple{AnSwER}}}}$

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.

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

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

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

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

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