Q. \[\int 5^x \, dx\]

Q: What is \int 5^x \, dx ?

\int 5^x \, dx = \dfrac{5^x}{\ln 5} + C .

Q: How do you integrate a^x in general?

\int a^x \, dx = \dfrac{a^x}{\ln a} + C for a>0,\ a\neq 1 .

Q: Can you show the substitution used to integrate 5^x ?

Let u=5^x. Then du=5^x\ln 5\,dx. So dx=\dfrac{du}{u\ln 5} and \int 5^x dx = \dfrac{1}{\ln 5}\int du = \dfrac{5^x}{\ln 5}+C.

Q: What is \int 5^{2x} \, dx ?

5^{2x}=(5^2)^x=25^x. Thus \int 5^{2x}\,dx = \dfrac{5^{2x}}{2\ln 5}+C .

Q: What is \int 5^{x} \ln(5)\, dx ?

Multiply the result by \ln 5: \int 5^{x}\ln(5)\,dx = \ln(5)\cdot \dfrac{5^x}{\ln 5}+C = 5^x + C .

Q: What is \int \dfrac{1}{5^x}\, dx ?

\dfrac{1}{5^x}=5^{-x}. So \int 5^{-x}\,dx = \dfrac{5^{-x}}{-\ln 5}+C = -\dfrac{5^{-x}}{\ln 5}+C .

Answer

To find \( \int 5^x \, dx \), use the rule

\[ \int a^x \, dx = \frac{a^x}{\ln a} + C \quad \text{for } a>0,\ a\ne 1. \]

Here \(a=5\). So

\[ \int 5^x \, dx = \frac{5^x}{\ln 5} + C. \]

Detailed Explanation

We want to find the indefinite integral

\[
\int 5^x \, dx.
\]

Step 1: Recognize the form

The function \(5^x\) is an exponential function with base \(5\). A standard rule is:

\[
\int a^x \, dx = \frac{a^x}{\ln(a)} + C \quad \text{for } a > 0,\ a \ne 1.
\]

Step 2: Apply the rule with \(a = 5\)

Here \(a^x = 5^x\), so \(a = 5\). Substitute into the formula:

\[
\int 5^x \, dx = \frac{5^x}{\ln(5)} + C.
\]

Final Answer

\[
\int 5^x \, dx = \frac{5^x}{\ln(5)} + C.
\]

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Calculus FAQ

What is \( \int 5^x \, dx \) ?

\( \int 5^x \, dx = \dfrac{5^x}{\ln 5} + C \).

How do you integrate \( a^x \) in general?

\( \int a^x \, dx = \dfrac{a^x}{\ln a} + C \) for \( a>0,\ a\neq 1 \).

Can you show the substitution used to integrate \(5^x\) ?

Let \(u=5^x\). Then \(du=5^x\ln 5\,dx\). So \(dx=\dfrac{du}{u\ln 5}\) and \( \int 5^x dx = \dfrac{1}{\ln 5}\int du = \dfrac{5^x}{\ln 5}+C\).

What is \( \int 5^{2x} \, dx \) ?

\(5^{2x}=(5^2)^x=25^x\). Thus \( \int 5^{2x}\,dx = \dfrac{5^{2x}}{2\ln 5}+C \).

What is \( \int 5^{x} \ln(5)\, dx \) ?

Multiply the result by \(\ln 5\): \( \int 5^{x}\ln(5)\,dx = \ln(5)\cdot \dfrac{5^x}{\ln 5}+C = 5^x + C \).

What is \( \int \dfrac{1}{5^x}\, dx \) ?

\( \dfrac{1}{5^x}=5^{-x}\). So \( \int 5^{-x}\,dx = \dfrac{5^{-x}}{-\ln 5}+C = -\dfrac{5^{-x}}{\ln 5}+C \).

Solve ∫5^x step by step.
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