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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= 7 \cdot \int \frac{1}{2 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}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 7 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

f (x) = sec (θ) tan (θ)

\int x^{2} (5 x^{2} + x)^{2} d x

\frac{\partial}{\partial h} (\frac{1}{2} (a + b) h)

\frac{d}{d x} (\frac{2}{x ^{\frac{1}{3}}} - 2)

f (x) = - 5 x (2 x - 3)^{4}

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