What is the integral of sqrt(10^{5x)} ?

The integral of sqrt(10^{5x)} is (2^{1/2 (2+5x)}*5^{1/2 (5x-2)})/(ln(10))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sqrt(10^{5x)}

\int 1 0^{5 x} d x

\frac{2 ^{\frac{1}{2} (2 + 5 x)} \cdot 5 ^{\frac{1}{2} (5 x - 2)}}{ln ( 10 )} + C

Solution steps

\int 1 0^{5 x} d x

Apply radical rule:

n a^{m} = a^{\frac{m}{n}},

assuming

a \geq 0

= \int 1 0^{\frac{5 x}{2}} d x

Apply u-substitution

: \int 2^{1 + u} \cdot 5^{u - 1} d u

= \int 2^{1 + u} \cdot 5^{u - 1} d u

Simplify

2^{1 + u} : 2 \cdot 2^{u}

= \int 2 \cdot 2^{u} \cdot 5^{u - 1} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int 2^{u} \cdot 5^{u - 1} d u

Simplify

5^{u - 1} : 5^{u} \cdot 5^{- 1}

= 2 \cdot \int 2^{u} \cdot 5^{u} \cdot 5^{- 1} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot 5^{- 1} \cdot \int 2^{u} \cdot 5^{u} d u

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= 2 \cdot \frac{1}{5} \cdot \int 2^{u} \cdot 5^{u} d u

Simplify

2^{u} \cdot 5^{u} : 1 0^{u}

= 2 \cdot \frac{1}{5} \cdot \int 1 0^{u} d u

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

= 2 \cdot \frac{1}{5} \cdot \frac{1 0 ^{u}}{ln ( 10 )}

Substitute back

u = \frac{5 x}{2}

= 2 \cdot \frac{1}{5} \cdot \frac{1 0 ^{\frac{5 x}{2}}}{ln ( 10 )}

Simplify

2 \cdot \frac{1}{5} \cdot \frac{1 0 ^{\frac{5 x}{2}}}{ln ( 10 )} : \frac{2 ^{\frac{1}{2} (2 + 5 x)} \cdot 5 ^{\frac{1}{2} (5 x - 2)}}{ln ( 10 )}

= \frac{2 ^{\frac{1}{2} (2 + 5 x)} \cdot 5 ^{\frac{1}{2} (5 x - 2)}}{ln ( 10 )}

Add a constant to the solution

= \frac{2 ^{\frac{1}{2} (2 + 5 x)} \cdot 5 ^{\frac{1}{2} (5 x - 2)}}{ln ( 10 )} + C

\int (- 2 x^{2} + 6 x) d x

\int e^{x l n (a)} d x

\int \frac{5 x + 3}{x ^{2} + 3 x + 4} d x

\int cos^{7} (θ) sin (2 θ) d θ

\int 2 x e^{x y} d x

What is the integral of sqrt(10^{5x)} ?
The integral of sqrt(10^{5x)} is (2^{1/2 (2+5x)}*5^{1/2 (5x-2)})/(ln(10))+C