What is the integral of x/(4x^2+1) ?

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

Popular >

integral of x/(4x^2+1)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{8 u} d u

= \int \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{8} ln ∣ u ∣

Substitute back

u = 4 x^{2} + 1

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

Add a constant to the solution

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

\int e^{\frac{t}{5}} d t

\int \frac{( x + 3 )}{x ^{2} + 8 x - 9} d x

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

n = 1 \sum \infty n^{3} (1 + n^{4})

f (x) = x \cdot e^{2 x - 4}, x = 2

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