What is the integral of (x^2+y^2-5)/((1+x)^3) ?

The integral of (x^2+y^2-5)/((1+x)^3) is ln|1+x|+2/(1+x)-(y^2)/(2(1+x)^2)+2/((1+x)^2)+C

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integral of (x^2+y^2-5)/((1+x)^3)

\int \frac{x ^{2} + y ^{2} - 5}{( 1 + x ) ^{3}} d x

ln ∣ 1 + x ∣ + \frac{2}{1 + x} - \frac{y ^{2}}{2 ( 1 + x ) ^{2}} + \frac{2}{( 1 + x ) ^{2}} + C

Solution steps

\int \frac{x ^{2} + y ^{2} - 5}{( 1 + x ) ^{3}} d x

Apply u-substitution

: \int \frac{u ^{2} - 2 u + y ^{2} - 4}{u ^{3}} d u

= \int \frac{u ^{2} - 2 u + y ^{2} - 4}{u ^{3}} d u

Expand

\frac{u ^{2} - 2 u + y ^{2} - 4}{u ^{3}} : \frac{1}{u} - \frac{2}{u ^{2}} + \frac{y ^{2}}{u ^{3}} - \frac{4}{u ^{3}}

= \int \frac{1}{u} - \frac{2}{u ^{2}} + \frac{y ^{2}}{u ^{3}} - \frac{4}{u ^{3}} 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 \frac{1}{u} d u - \int \frac{2}{u ^{2}} d u + \int \frac{y ^{2}}{u ^{3}} d u - \int \frac{4}{u ^{3}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

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

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

\int \frac{4}{u ^{3}} d u = - \frac{2}{u ^{2}}

= ln ∣ u ∣ - (- \frac{2}{u}) - \frac{y ^{2}}{2 u ^{2}} - (- \frac{2}{u ^{2}})

Substitute back

u = 1 + x

= ln ∣ 1 + x ∣ - (- \frac{2}{1 + x}) - \frac{y ^{2}}{2 ( 1 + x ) ^{2}} - (- \frac{2}{( 1 + x ) ^{2}})

Simplify

= ln ∣ 1 + x ∣ + \frac{2}{1 + x} - \frac{y ^{2}}{2 ( 1 + x ) ^{2}} + \frac{2}{( 1 + x ) ^{2}}

Add a constant to the solution

= ln ∣ 1 + x ∣ + \frac{2}{1 + x} - \frac{y ^{2}}{2 ( 1 + x ) ^{2}} + \frac{2}{( 1 + x ) ^{2}} + C

\frac{d}{d x} (a^{2} + x^{2})

\frac{d v}{d t} = \frac{t}{sin ( v )}

\frac{d}{d x} (e^{x} (6 x^{2} - 12 x + 12))

\int \frac{( sin ( x ) ) ^{2}}{x} d x

\int_{1}^{5} \frac{3}{x} d x

What is the integral of (x^2+y^2-5)/((1+x)^3) ?
The integral of (x^2+y^2-5)/((1+x)^3) is ln|1+x|+2/(1+x)-(y^2)/(2(1+x)^2)+2/((1+x)^2)+C