What is the integral of x/((ax+b)^4) ?

The integral of x/((ax+b)^4) is (-3ax-b)/(6a^2(ax+b)^3)+C

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integral of x/((ax+b)^4)

\int \frac{x}{( a x + b ) ^{4}} d x

\frac{- 3 a x - b}{6 a ^{2} ( a x + b ) ^{3}} + C

Solution steps

\int \frac{x}{( a x + b ) ^{4}} d x

Apply u-substitution

= \int \frac{u - b}{a ^{2} u ^{4}} d u

Take the constant out:

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

= \frac{1}{a ^{2}} \cdot \int \frac{u - b}{u ^{4}} d u

Expand

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

= \frac{1}{a ^{2}} \cdot \int \frac{1}{u ^{3}} - \frac{b}{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

= \frac{1}{a ^{2}} (\int \frac{1}{u ^{3}} d u - \int \frac{b}{u ^{4}} d u)

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

\int \frac{b}{u ^{4}} d u = - \frac{b}{3 u ^{3}}

= \frac{1}{a ^{2}} (- \frac{1}{2 u ^{2}} - (- \frac{b}{3 u ^{3}}))

Substitute back

u = a x + b

= \frac{1}{a ^{2}} (- \frac{1}{2 ( a x + b ) ^{2}} - (- \frac{b}{3 ( a x + b ) ^{3}}))

Simplify

\frac{1}{a ^{2}} (- \frac{1}{2 ( a x + b ) ^{2}} - (- \frac{b}{3 ( a x + b ) ^{3}})) : \frac{- 3 a x - b}{6 a ^{2} ( a x + b ) ^{3}}

= \frac{- 3 a x - b}{6 a ^{2} ( a x + b ) ^{3}}

Add a constant to the solution

= \frac{- 3 a x - b}{6 a ^{2} ( a x + b ) ^{3}} + C

\int - \frac{2}{3} d x

\int \frac{1}{2 x ^{2} - 3} d x

\int \frac{2 + 6 x}{1 + x ^{2}} d x

\int \frac{1}{4 + 5 sin ^{2} ( x )} d x

\int 3 \cdot \frac{ln ( x )}{x ^{3}} d x

What is the integral of x/((ax+b)^4) ?
The integral of x/((ax+b)^4) is (-3ax-b)/(6a^2(ax+b)^3)+C