What is the integral of 1/(x^2+xsqrt(x)) ?

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

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integral of 1/(x^2+xsqrt(x))

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2}{u ^{2} ( u + 1 )} d u

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{u ^{2} ( u + 1 )} : - \frac{1}{u} + \frac{1}{u ^{2}} + \frac{1}{u + 1}

= 2 \cdot \int - \frac{1}{u} + \frac{1}{u ^{2}} + \frac{1}{u + 1} 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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

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

Substitute back

u = x

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

Add a constant to the solution

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

x \to 7 lim (2)

f (x) = \frac{5}{x}, at x = 3

\int 36 x (9 x^{2} - 7)^{3} d x

\int 5 tan (5 x) d x

x \to 4 - lim (x + 1)

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