What is the integral of (x^2)/((x+a)) ?

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

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integral of (x^2)/((x+a))

integral

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

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

Solution steps

\int \frac{x ^{2}}{x + a} d x

x + a = a (\frac{x}{a} + 1)

= \int \frac{x ^{2}}{a ( \frac{x}{a} + 1 )} d x

Take the constant out:

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

= \frac{1}{a} \cdot \int \frac{x ^{2}}{\frac{x}{a} + 1} d x

Apply u-substitution

: \int \frac{a ( a u - a ) ^{2}}{u} d u

= \frac{1}{a} \cdot \int \frac{a ( a u - a ) ^{2}}{u} d u

Take the constant out:

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

= \frac{1}{a} a \cdot \int \frac{( a u - a ) ^{2}}{u} d u

Expand

\frac{( a u - a ) ^{2}}{u} : a^{2} u - 2 a^{2} + \frac{a ^{2}}{u}

= \frac{1}{a} a \cdot \int a^{2} u - 2 a^{2} + \frac{a ^{2}}{u} 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} a (\int a^{2} u d u - \int 2 a^{2} d u + \int \frac{a ^{2}}{u} d u)

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

\int 2 a^{2} d u = 2 a^{2} u

\int \frac{a ^{2}}{u} d u = a^{2} ln ∣ u ∣

= \frac{1}{a} a (\frac{a ^{2} u ^{2}}{2} - 2 a^{2} u + a^{2} ln ∣ u ∣)

Substitute back

u = \frac{x}{a} + 1

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

Simplify

\frac{1}{a} a (\frac{a ^{2} ( \frac{x}{a} + 1 ) ^{2}}{2} - 2 a^{2} (\frac{x}{a} + 1) + a^{2} ln \frac{x}{a} + 1) : \frac{1}{2} (x + a)^{2} - 2 a^{2} (\frac{x}{a} + 1) + a^{2} ln \frac{x}{a} + 1

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

Add a constant to the solution

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

- 6 x^{3} + 9 x^{2} + 4 x - 3

e^{4} x

1 - f (x)

3 x^{2} + 4 x

cos^{2} (3 x)

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