What is the integral from 0 to x of 3/((1+x)^4) ?

The integral from 0 to x of 3/((1+x)^4) is -1/((1+x)^3)+1

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integral from 0 to x of 3/((1+x)^4)

\int_{0}^{x} \frac{3}{( 1 + x ) ^{4}} d x

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

Solution steps

\int_{0}^{x} \frac{3}{( 1 + x ) ^{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}^{x} \frac{1}{( 1 + x ) ^{4}} d x

Apply u-substitution

= 3 \cdot \int_{1}^{1 + x} \frac{1}{u ^{4}} d u

Apply the Power Rule

= 3 [- \frac{1}{3 u ^{3}}]_{1}^{1 + x}

Compute the boundaries:

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

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

Simplify

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

What is the integral from 0 to x of 3/((1+x)^4) ?
The integral from 0 to x of 3/((1+x)^4) is -1/((1+x)^3)+1