Answered: "Evaluate the integral. ∫2xxdx"

aidmoong7

Answered question

2021-11-16

Evaluate the integral.

\int \frac{2^{\sqrt{x}}}{\sqrt{x}} d x

Anthony Caraballo

Beginner2021-11-17Added 15 answers

Step 1
Integral:

\int \frac{2^{\sqrt{x}}}{\sqrt{x}} d x

Let,

\sqrt{x} = z

\frac{1}{2 \sqrt{x}} d x = d z

Step 2

= \int 2^{z + 1} d z

= 2 \cdot \frac{2^{z}}{\ln 2} + c

= 2 \cdot \frac{2^{\sqrt{x}}}{\ln 2} + c

This is the required integration.

Vaing1990

Beginner2021-11-18Added 16 answers

Step 1: Use Power Substitution.
Let

u = \sqrt{x}, = u^{2}

, and dx=2u du
Step 2: Substitute variables from above.

\int \frac{2^{u}}{u} \times 2 u d u

Step 3: Use Integration by Substitution.
Let w=u+1, dw=du
Step 4: Using w and dw above, rewrite

\int 2^{u + 1} d u

\int 2^{w} d w

Step 5: Use this property:

\int a^{x} d x = \frac{a^{x}}{\ln a}

\frac{2^{w}}{\ln 2}

Step 6: Substitute w=u+1 back into the original integral.

\frac{2^{u + 1}}{\ln 2}

Step 7: Substitute

u = \sqrt{x}

back into the original integral.

\frac{2^{\sqrt{x} + 1}}{\ln 2}

Step 8: Add constant.

\frac{2^{\sqrt{x} + 1}}{\ln 2} + C

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