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

**Integration of 2**^{x} Formula

^{x}Formula

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

∫ 2^{x} dx = 2^{x}/ln2 +C.

**How to Integrate 2**^{x}

^{x}

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

*Proof:*

Note that x can be written as x=e^{lnx}, so we have

2=e^{ln2} **…(I)**

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

∫2^{x} dx = ∫(e^{ln2})^{x} dx

⇒ ∫2^{x} dx = ∫e^{xln2} dx **…(II)**

Let xln2 =z. Then ln2 dx =dz

⇒ dx = dz/ln2

Now, from from **(II)** we get that

∫2^{x} dx = ∫e^{z} $\dfrac{dz}{\ln 2}$

= $\dfrac{1}{\ln 2}$ ∫e^{z} dz

= e^{z}/ln2 as the integration of e^{z} with respect to z is e^{z}.

= e^{xln2}/ln2 as z=xln2.

= 2^{x}/ln2 using the identity **(I)** as e^{xln2} = (e^{ln2})^{x} =(2)^{x} = 2^{x}.

So the integration of 2^x is equal to 2^{x}/ln2 where ln2=log_{e}2. This is obtained by the substitution method of integration and using the exponential identity x=e^{lnx}.

**Video Solution of Integration of 2^x:**

**ALSO READ:**

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

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

**FAQs**

**Q1: What is the Integration of 2^x?**

**Answer:** The integration of 2^x is equal to 2^{x}/ln2 where ln2=log_{e}2.