What is the integral of (7x)/(36x^2+1) ?

The integral of (7x)/(36x^2+1) is 7/72 ln|36x^2+1|+C

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integral of (7x)/(36x^2+1)

\int \frac{7 x}{36 x ^{2} + 1} d x

\frac{7}{72} ln 36 x^{2} + 1 + C

Solution steps

\int \frac{7 x}{36 x ^{2} + 1} d x

Take the constant out:

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

= 7 \cdot \int \frac{x}{36 x ^{2} + 1} d x

Apply u-substitution

: \int \frac{1}{72 u} d u

= 7 \cdot \int \frac{1}{72 u} d u

Take the constant out:

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

= 7 \cdot \frac{1}{72} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 7 \cdot \frac{1}{72} ln ∣ u ∣

Substitute back

u = 36 x^{2} + 1

= 7 \cdot \frac{1}{72} ln 36 x^{2} + 1

Simplify

7 \cdot \frac{1}{72} ln 36 x^{2} + 1 : \frac{7}{72} ln 36 x^{2} + 1

= \frac{7}{72} ln 36 x^{2} + 1

Add a constant to the solution

= \frac{7}{72} ln 36 x^{2} + 1 + C

x = 10 - 2 y, x = y^{2} + 2

f^{^{'}} (x) = x + 4

\int \frac{( x ^{3} - 1 )}{x - 1} d x

\int_{\frac{π}{2}}^{π} x sin^{2} (x) d x

\frac{d}{d x} (\frac{1 - y}{x - 1})

What is the integral of (7x)/(36x^2+1) ?
The integral of (7x)/(36x^2+1) is 7/72 ln|36x^2+1|+C