Question

evaluate int. x log dx​

Answer 1

Appropriate Question :- Evaluate

$\rm \: \displaystyle\int\rm x \: logx \: dx$

$\red{\large\underline{\sf{Solution-}}}$

Given integral is

$\rm \: \displaystyle\int\rm x \: logx \: dx$

By using, Integration by Parts, we have

$\rm \: = \: logx\displaystyle\int\rm x \: dx \: - \: \displaystyle\int\rm \bigg[\dfrac{d}{dx}logx\displaystyle\int\rm x \: dx \bigg]dx$

$\rm \: = \: logx \times \dfrac{ {x}^{2} }{2} - \displaystyle\int\rm \dfrac{1}{x} \times \frac{ {x}^{2} }{2} \: dx$

$\rm \: = \: \dfrac{ {x}^{2} logx}{2} - \dfrac{1}{2}\displaystyle\int\rm x \: dx$

$\rm \: = \: \dfrac{ {x}^{2} logx}{2} - \dfrac{1}{2} \times \dfrac{ {x}^{2} }{2} + c$

$\rm \: = \: \dfrac{ {x}^{2} logx}{2} - \dfrac{ {x}^{2} }{4} + c$

Hence,

$\rm\implies \:\boxed{ \rm{ \: \displaystyle\int\rm x \: logx \: dx = \: \dfrac{ {x}^{2} logx}{2} - \dfrac{ {x}^{2} }{4} + c \: }}$

$\rule{190pt}{2pt}$

Basic Concept Used :-

Integration by Parts

$\boxed{ \rm{ \:\displaystyle\int\rm (u.v)dx = u\displaystyle\int\rm v \: dx - \displaystyle\int\rm \bigg[\dfrac{d}{dx}u\displaystyle\int\rm vdx \bigg]dx \: }}$

Where,

• u is the function u(x)

• v is the function v(x)

• u' is the derivative of the function u(x)

The function u and v are chosen according to the word ILATE

where,

• I - Inverse trigonometric functions

• L - Logarithmic functions

• A - Algebraic functions

• T - Trigonometric functions

• E- Exponential functions

The alphabet which comes first is taken as u and other as v.

$\rule{190pt}{2pt}$

Additional information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

Appropriate Question :- Evaluate

\begin{gathered}\rm \: \displaystyle\int\rm x \: logx \: dx \\ \end{gathered}

∫xlogxdx

\red{\large\underline{\sf{Solution-}}}

Solution−

Given integral is

\begin{gathered}\rm \: \displaystyle\int\rm x \: logx \: dx \\ \end{gathered}

∫xlogxdx

By using, Integration by Parts, we have

\rm \: = \: logx\displaystyle\int\rm x \: dx \: - \: \displaystyle\int\rm \bigg[\dfrac{d}{dx}logx\displaystyle\int\rm x \: dx \bigg]dx=logx∫xdx−∫[

dx

d

logx∫xdx]dx

\rm \: = \: logx \times \dfrac{ {x}^{2} }{2} - \displaystyle\int\rm \dfrac{1}{x} \times \frac{ {x}^{2} }{2} \: dx=logx×

2

x

2

−∫

x

1

×

2

x

2

dx

\rm \: = \: \dfrac{ {x}^{2} logx}{2} - \dfrac{1}{2}\displaystyle\int\rm x \: dx=

2

x

2

logx

−

2

1

∫xdx

\begin{gathered}\rm \: = \: \dfrac{ {x}^{2} logx}{2} - \dfrac{1}{2} \times \dfrac{ {x}^{2} }{2} + c \\ \end{gathered}

=

2

x

2

logx

−

2

1

×

2

x

2

+c

\begin{gathered}\rm \: = \: \dfrac{ {x}^{2} logx}{2} - \dfrac{ {x}^{2} }{4} + c \\ \end{gathered}

=

2

x

2

logx

−

4

x

2

+c

Hence,

\begin{gathered}\rm\implies \:\boxed{ \rm{ \: \displaystyle\int\rm x \: logx \: dx = \: \dfrac{ {x}^{2} logx}{2} - \dfrac{ {x}^{2} }{4} + c \: }} \\ \end{gathered}

⟹

∫xlogxdx=

2

x

2

logx

−

4

x

2

+c

\rule{190pt}{2pt}

Basic Concept Used :-

Integration by Parts

\begin{gathered}\boxed{ \rm{ \:\displaystyle\int\rm (u.v)dx = u\displaystyle\int\rm v \: dx - \displaystyle\int\rm \bigg[\dfrac{d}{dx}u\displaystyle\int\rm vdx \bigg]dx \: }} \\ \end{gathered}

∫(u.v)dx=u∫vdx−∫[

dx

d

u∫vdx]dx

Where,

u is the function u(x)

v is the function v(x)

u' is the derivative of the function u(x)

The function u and v are chosen according to the word ILATE

where,

I - Inverse trigonometric functions

L - Logarithmic functions

A - Algebraic functions

T - Trigonometric functions

E- Exponential functions

The alphabet which comes first is taken as u and other as v.

\rule{190pt}{2pt}

Additional information :-

\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}\end{gathered}

f(x)

k

sinx

cosx

sec

2

x

cosec

2

x

secxtanx

cosecxcotx

tanx

x

1

e

x

∫f(x)dx

kx+c

−cosx+c

sinx+c

tanx+c

−cotx+c

secx+c

−cosecx+c

logsecx+c

logx+c

e

x

+c

$answer}$