What is the integral of (3(x+2))/((x+7)^4) ?

The integral of (3(x+2))/((x+7)^4) is -3/(2(x+7)^2)+5/((x+7)^3)+C

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integral of (3(x+2))/((x+7)^4)

\int \frac{3 ( x + 2 )}{( x + 7 ) ^{4}} d x

- \frac{3}{2 ( x + 7 ) ^{2}} + \frac{5}{( x + 7 ) ^{3}} + C

Solution steps

\int \frac{3 ( x + 2 )}{( x + 7 ) ^{4}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x + 2}{( x + 7 ) ^{4}} d x

Apply u-substitution

: \int \frac{u}{( u + 5 ) ^{4}} d u

= 3 \cdot \int \frac{u}{( u + 5 ) ^{4}} d u

Apply u-substitution

: \int \frac{v - 5}{v ^{4}} d v

= 3 \cdot \int \frac{v - 5}{v ^{4}} d v

Expand

\frac{v - 5}{v ^{4}} : \frac{1}{v ^{3}} - \frac{5}{v ^{4}}

= 3 \cdot \int \frac{1}{v ^{3}} - \frac{5}{v ^{4}} d v

Apply the Sum Rule:

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

= 3 (\int \frac{1}{v ^{3}} d v - \int \frac{5}{v ^{4}} d v)

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

\int \frac{5}{v ^{4}} d v = - \frac{5}{3 v ^{3}}

= 3 (- \frac{1}{2 v ^{2}} - (- \frac{5}{3 v ^{3}}))

Substitute back

= 3 (- \frac{1}{2 ( x + 2 + 5 ) ^{2}} - (- \frac{5}{3 ( x + 2 + 5 ) ^{3}}))

Simplify

3 (- \frac{1}{2 ( x + 2 + 5 ) ^{2}} - (- \frac{5}{3 ( x + 2 + 5 ) ^{3}})) : - \frac{3}{2 ( x + 7 ) ^{2}} + \frac{5}{( x + 7 ) ^{3}}

= - \frac{3}{2 ( x + 7 ) ^{2}} + \frac{5}{( x + 7 ) ^{3}}

Add a constant to the solution

= - \frac{3}{2 ( x + 7 ) ^{2}} + \frac{5}{( x + 7 ) ^{3}} + C

\frac{d}{d x} (cos^{2} (π x))

\int_{0}^{1} - \frac{5 y}{e ^{2 y}} d y

- x^{3} + x, x = - 1, y = - \frac{1}{2}

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

\int_{0}^{5} \frac{2}{x ^{2}} d x

What is the integral of (3(x+2))/((x+7)^4) ?
The integral of (3(x+2))/((x+7)^4) is -3/(2(x+7)^2)+5/((x+7)^3)+C