What is the integral from-0.9 to-0.1 of (3e^{3x})/((e^{3x)+3)^4} ?

The integral from-0.9 to-0.1 of (3e^{3x})/((e^{3x)+3)^4} is 0.00518…

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integral from-0.9 to-0.1 of (3e^{3x})/((e^{3x)+3)^4}

\int_{- 0.9}^{- 0.1} \frac{3 e ^{3 x}}{( e ^{3 x} + 3 ) ^{4}} d x

0.00518 \dots

Solution steps

\int_{- 0.9}^{- 0.1} \frac{3 e ^{3 x}}{( e ^{3 x} + 3 ) ^{4}} d x

Take the constant out:

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

= 3 \cdot \int_{- 0.9}^{- 0.1} \frac{e ^{3 x}}{( e ^{3 x} + 3 ) ^{4}} d x

Apply u-substitution

: \int_{- 2.7}^{- 0.3} \frac{e ^{u}}{3 ( e ^{u} + 3 ) ^{4}} d u

= 3 \cdot \int_{- 2.7}^{- 0.3} \frac{e ^{u}}{3 ( e ^{u} + 3 ) ^{4}} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int_{- 2.7}^{- 0.3} \frac{e ^{u}}{( e ^{u} + 3 ) ^{4}} d u

Apply u-substitution

: \int_{3.06720 \dots}^{3.74081 \dots} \frac{1}{v ^{4}} d v

= 3 \cdot \frac{1}{3} \cdot \int_{3.06720 \dots}^{3.74081 \dots} \frac{1}{v ^{4}} d v

Apply the Power Rule

: [- \frac{1}{3 v ^{3}}]_{3.06720 \dots}^{3.74081 \dots}

= 3 \cdot \frac{1}{3} [- \frac{1}{3 v ^{3}}]_{3.06720 \dots}^{3.74081 \dots}

Simplify

3 \cdot \frac{1}{3} [- \frac{1}{3 v ^{3}}]_{3.06720 \dots}^{3.74081 \dots} : [- \frac{0.33333 \dots}{v ^{3}}]_{3.06720 \dots}^{3.74081 \dots}

= [- \frac{0.33333 \dots}{v ^{3}}]_{3.06720 \dots}^{3.74081 \dots}

Compute the boundaries:

0.00518 \dots

= 0.00518 \dots

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What is the integral from-0.9 to-0.1 of (3e^{3x})/((e^{3x)+3)^4} ?
The integral from-0.9 to-0.1 of (3e^{3x})/((e^{3x)+3)^4} is 0.00518…