What is the integral from 0 to 9 of x/((x+9)^4) ?

The integral from 0 to 9 of x/((x+9)^4) is 1/972

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integral from 0 to 9 of x/((x+9)^4)

\int_{0}^{9} \frac{x}{( x + 9 ) ^{4}} d x

\frac{1}{972}

Decimal Notation

0.00102 \dots

Solution steps

\int_{0}^{9} \frac{x}{( x + 9 ) ^{4}} d x

Apply u-substitution

: \int_{9}^{18} \frac{u - 9}{u ^{4}} d u

= \int_{9}^{18} \frac{u - 9}{u ^{4}} d u

Expand

\frac{u - 9}{u ^{4}} : \frac{1}{u ^{3}} - \frac{9}{u ^{4}}

= \int_{9}^{18} \frac{1}{u ^{3}} - \frac{9}{u ^{4}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{9}^{18} \frac{1}{u ^{3}} d u - \int_{9}^{18} \frac{9}{u ^{4}} d u

\int_{9}^{18} \frac{1}{u ^{3}} d u = \frac{1}{216}

\int_{9}^{18} \frac{9}{u ^{4}} d u = \frac{7}{1944}

= \frac{1}{216} - \frac{7}{1944}

Simplify

\frac{1}{216} - \frac{7}{1944} : \frac{1}{972}

= \frac{1}{972}

\int_{1}^{2} (x - 2)^{- 2} d x

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\int_{0}^{1} sin (π x cos (π) x) d x

\int 4 x^{3} + 1 d x

What is the integral from 0 to 9 of x/((x+9)^4) ?
The integral from 0 to 9 of x/((x+9)^4) is 1/972