Integration of 2^x | 2^x Integration

The integration of 2x is equal to 2x/ln2 where ln denotes the natural logarithm, that is, ln=loge. In this post, we will learn to integrate 2 to the power x with respect to x.

Integral of 2^x

Integration of 2x Formula

The integration formula of 2x (2 to the x) with respect to x is given below.

∫ 2x dx = 2x/ln2 +C.

How to Integrate 2x

Now, we will learn how to integrate the function 2 to the x and prove that ∫ 2x dx = 2x/ln2 +C.


Note that x can be written as x=elnx, so we have

2=eln2 …(I)

Then, using this identity, the integration of 2^x will be equal to

∫2x dx = ∫(eln2)x dx

⇒ ∫2x dx = ∫exln2 dx …(II)

Let xln2 =z. Then ln2 dx =dz

⇒ dx = dz/ln2

Now, from from (II) we get that

∫2x dx = ∫ez $\dfrac{dz}{\ln 2}$

⇒ ∫2x dx = $\dfrac{1}{\ln 2}$ ∫ez dz

⇒ ∫2x dx = ez/ln2 as the integration of ez with respect to z is ez.

⇒ ∫2x dx = exln2/ln2 as z=xln2.

⇒ ∫2x dx = 2x/ln2
[Here we have used the identity (I) as exln2 = (eln2)x =(2)x = 2x.]

So the integration of 2^x is equal to 2x/ln2 where ln2=loge2. This is obtained by the substitution method of integration and using the exponential identity x=elnx.

Video Solution of Integration of 2^x:


Question: Find the definite integral ∫01 2x dx.


Using the above integration formula of 2x, it follows that

01 2x dx = $\Big[ \dfrac{2^x}{\ln 2}\Big]$

⇒ ∫01 2x dx = $\dfrac{2^1}{\ln 2} – \dfrac{2^0}{\ln 2}$

⇒ ∫01 2x dx = $\dfrac{1}{\ln 2} (2-1)$

⇒ ∫01 2x dx = 1/ln2.

So the definite integral of 2x from 0 to 1 is equal to ln2.


Integration of 1/(1+x2)

Integration of log(sin x)

Integration of $\sqrt{a^2-x^2}$

Integration of $\sqrt{a^2+x^2}$


Q1: What is the Integration of 2^x?

Answer: The integration of 2^x is equal to 2x/ln2 where ln2=loge2.

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