What is the integral of (3e^x)/(e^{2x)+10e^x+25} ?

The integral of (3e^x)/(e^{2x)+10e^x+25} is -3/(e^x+5)+C

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integral of (3e^x)/(e^{2x)+10e^x+25}

\int \frac{3 e ^{x}}{e ^{2 x} + 10 e ^{x} + 25} d x

- \frac{3}{e ^{x} + 5} + C

Solution steps

\int \frac{3 e ^{x}}{e ^{2 x} + 10 e ^{x} + 25} d x

Take the constant out:

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

= 3 \cdot \int \frac{e ^{x}}{e ^{2 x} + 10 e ^{x} + 25} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 10 u + 25} d u

= 3 \cdot \int \frac{1}{u ^{2} + 10 u + 25} d u

Take the partial fraction of

\frac{1}{u ^{2} + 10 u + 25} : \frac{1}{( u + 5 ) ^{2}}

= 3 \cdot \int \frac{1}{( u + 5 ) ^{2}} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= 3 \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 3 (- \frac{1}{v})

Substitute back

= 3 (- \frac{1}{e ^{x} + 5})

Simplify

3 (- \frac{1}{e ^{x} + 5}) : - \frac{3}{e ^{x} + 5}

= - \frac{3}{e ^{x} + 5}

Add a constant to the solution

= - \frac{3}{e ^{x} + 5} + C

\int \frac{( e ^{x} + 9 )}{( e ^{2 x} - 6 e ^{x} + 9 )} d x

f (x) = \frac{8}{x}, at x = 4

y = 2 x, 2 x - 4, y = 0

5 x^{3} - 10 x^{2} - \frac{3}{x}

\frac{d}{d x} (sin (x^{3} + 2 x - 1))

What is the integral of (3e^x)/(e^{2x)+10e^x+25} ?
The integral of (3e^x)/(e^{2x)+10e^x+25} is -3/(e^x+5)+C