Question

integrate (2xsquare+x)dx​

Answer 1

Solution:-

We have

$\to \sf\int(2 {x}^{2} + x)dx$

Now we using This Identity

$\sf \to \: \int {x}^{n} dx = \dfrac{x {}^{n + 1} }{n + 1} + c \: \: \: where \: n \not = \: - 1$

Now

$\to \sf\int(2 {x}^{2} + x)dx$

$\sf \to \: \int2 {x}^{2} dx + \int{x}dx$

$\sf \to \: 2 \int {x}^{2} dx + \int{x}dx$

$\sf \to \: 2 \bigg(\dfrac{ {x}^{2 + 1} }{2 + 1} \bigg) + \dfrac{ {x}^{1 + 1} }{1 + 1} + c$

$\sf \to \: 2 \bigg( \dfrac{ {x}^{3} }{3} \bigg) + \dfrac{ {x}^{2} }{2} + c$

$\sf \to \: \dfrac{2 {x}^{3} }{3} + \dfrac{ {x}^{2} }{2} + c$

Some Formula of Integration

$\sf \to \: \int {x}^{n} dx = \dfrac{x {}^{n + 1} }{n + 1} + c \: \: \: where \: n \not = \: - 1$

$\sf \to \: \int \dfrac{1}{x}dx = log_{e} |x| + c$

$\sf \to \: \int {e}^{x} dx = {e}^{x} + c$

$\sf \to \int {a}^{x} dx = \dfrac{ {a}^{x} }{ log_{e}a } + c$

Answer 2

Answer:

We have

\begin{gathered} \to \sf\int(2 {x}^{2} + x)dx \\ \end{gathered}

→∫(2x

2

+x)dx

Now we using This Identity

\begin{gathered} \sf \to \: \int {x}^{n} dx = \dfrac{x {}^{n + 1} }{n + 1} + c \: \: \: where \: n \not = \: - 1 \\ \end{gathered}

→∫x

n

dx=

n+1

x

n+1

+cwheren



=−1

Now

\begin{gathered} \to \sf\int(2 {x}^{2} + x)dx \\ \end{gathered}

→∫(2x

2

+x)dx

\begin{gathered} \sf \to \: \int2 {x}^{2} dx + \int{x}dx \\ \end{gathered}

→∫2x

2

dx+∫xdx

\begin{gathered} \sf \to \: 2 \int {x}^{2} dx + \int{x}dx \\ \end{gathered}

→2∫x

2

dx+∫xdx

\sf \to \: 2 \bigg(\dfrac{ {x}^{2 + 1} }{2 + 1} \bigg) + \dfrac{ {x}^{1 + 1} }{1 + 1} + c→2(

2+1

x

2+1

)+

1+1

x

1+1

+c

\sf \to \: 2 \bigg( \dfrac{ {x}^{3} }{3} \bigg) + \dfrac{ {x}^{2} }{2} + c→2(

3

x

3

)+

2

x

2

+c

\sf \to \: \dfrac{2 {x}^{3} }{3} + \dfrac{ {x}^{2} }{2} + c→

3

2x

3

+

2

x

2

+c

Some Formula of Integration

\begin{gathered}\sf \to \: \int {x}^{n} dx = \dfrac{x {}^{n + 1} }{n + 1} + c \: \: \: where \: n \not = \: - 1 \\ \end{gathered}

→∫x

n

dx=

n+1

x

n+1

+cwheren



=−1

\begin{gathered} \sf \to \: \int \dfrac{1}{x}dx = log_{e} |x| + c \\ \end{gathered}

→∫

x

1

dx=log

e

∣x∣+c

\begin{gathered} \sf \to \: \int {e}^{x} dx = {e}^{x} + c \\ \end{gathered}

→∫e

x

dx=e

x

+c

\begin{gathered} \sf \to \int {a}^{x} dx = \dfrac{ {a}^{x} }{ log_{e}a } + c \\ \end{gathered}

→∫a

x

dx=

log

e

a

a

x

+c