What is the integral of x/(sqrt(x^2+8x)) ?

The integral of x/(sqrt(x^2+8x)) is sqrt(x^2+8x)-4ln(1/4 |sqrt(x^2+8x)+x+4|)+C

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integral of x/(sqrt(x^2+8x))

\int \frac{x}{x ^{2} + 8 x} d x

x^{2} + 8 x - 4 ln (\frac{1}{4} x^{2} + 8 x + x + 4) + C

Solution steps

\int \frac{x}{x ^{2} + 8 x} d x

Complete the square

x^{2} + 8 x : (x + 4)^{2} - 16

= \int \frac{x}{( x + 4 ) ^{2} - 16} d x

Apply u-substitution

: \int \frac{u - 4}{u ^{2} - 16} d u

= \int \frac{u - 4}{u ^{2} - 16} d u

Expand

\frac{u - 4}{u ^{2} - 16} : \frac{u}{u ^{2} - 16} - \frac{4}{u ^{2} - 16}

= \int \frac{u}{u ^{2} - 16} - \frac{4}{u ^{2} - 16} 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{u}{u ^{2} - 16} d u - \int \frac{4}{u ^{2} - 16} d u

\int \frac{u}{u ^{2} - 16} d u = u^{2} - 16

\int \frac{4}{u ^{2} - 16} d u = 4 ln (\frac{1}{4} u^{2} - 16 + u)

= u^{2} - 16 - 4 ln (\frac{1}{4} u^{2} - 16 + u)

Substitute back

u = x + 4

= (x + 4)^{2} - 16 - 4 ln (\frac{1}{4} (x + 4)^{2} - 16 + x + 4)

Simplify

(x + 4)^{2} - 16 - 4 ln (\frac{1}{4} (x + 4)^{2} - 16 + x + 4) : x^{2} + 8 x - 4 ln (\frac{1}{4} x^{2} + 8 x + x + 4)

= x^{2} + 8 x - 4 ln (\frac{1}{4} x^{2} + 8 x + x + 4)

Add a constant to the solution

= x^{2} + 8 x - 4 ln (\frac{1}{4} x^{2} + 8 x + x + 4) + C

y = a x^{3}

\int_{0}^{1} \frac{ln ( t )}{t} d t

\int + x^{3} e^{3 x^{2}} d x

x \to 1 + lim (\frac{x ^{2} - 4}{x - 1})

\int \frac{20000}{( x + 100 ) ^{3}} d x

What is the integral of x/(sqrt(x^2+8x)) ?
The integral of x/(sqrt(x^2+8x)) is sqrt(x^2+8x)-4ln(1/4 |sqrt(x^2+8x)+x+4|)+C