If - 22x- 2x d x-A -22x-c- then A-

The correct option is D 1(log 2)^2I=∫ 2^2^x.2^x dx 2^x=u ⇒ 2^x log 2 dx=du u log 2 dx=du =∫2^u · uu log 2 du =1log 2∫ 2^u du nbsp; 2^u=t ⇒ 2^u log 2 du= ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Theorems on Integration

If ∫ 22x· 2x ...

Question

If $\int 2^{2^{x}} {.2}^{x} d x = A {.2}^{2^{x}} + c$ , then $A =$

\frac{1}{log 2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

log 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(log 2)^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{(log 2)^{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{1}{(log 2)^{2}}$
$I = \int 2^{2^{x}} {.2}^{x} d x$
$2^{x} = u \Rightarrow 2^{x} log 2 d x = d u$
$u log 2 d x = d u$
$= \int \frac{2^{u} \cdot u}{u log 2} d u$
$= \frac{1}{log 2} \int 2^{u} d u$
$2^{u} = t \Rightarrow 2^{u} log 2 d u = d t$
$= \frac{1}{(log 2)^{2}} \int \frac{t}{t} d t$
$= \frac{t}{(log 2)^{2}} + c$
$= \frac{2^{u}}{(log 2)^{2}} + c$
$= \frac{2^{2^{x}}}{(log 2)^{2}} + c$

Suggest Corrections

Similar questions

Q. Evaluate :

\int 2^{2^{2^{x}}} 2^{2^{x}} 2^{x} d x .

Q. Evaluate

\int 2^{2^{2^{x}}} {.2}^{2^{x}} {.2}^{x} d x

$If cos (\frac{x}{2}) c o s (\frac{x}{2^{2}}) c o s (\frac{x}{2^{3}}) . . . . . . . . . . . . . . . t o \infty = \frac{s i n x}{x} t h e n \frac{1}{2^{2}} s e c^{2} (\frac{x}{2}) \frac{1}{2^{4}} s e c^{2} (\frac{x}{2^{2}}) + . . . . . . . =$

Q. Expand

(x - 3) (x - 19)

Q. If

y = 2^{2^{x}}

, then

\frac{d y}{d x} = ?

Join BYJU'S Learning Program

If ∫22x.2x dx=A.22x+c, then A=

The correct option is D 1(log2)2I=∫22x.2x dx 2x=u⇒2xlog2 dx=du ulog2 dx=du =∫2u⋅uulog2 du =1log2∫2udu 2u=t⇒2ulog2 du=dt =1(log2)2∫ttdt =t(log2)2+c =2u(log2)2+c =22x(log2)2+c

If $\int 2^{2^{x}} {.2}^{x} d x = A {.2}^{2^{x}} + c$ , then $A =$