Question

integration of x4/(1+x)dx

Answer 1

HELLO FRIEND HERE IS YOUR ANSWER,,,,,

Given,,,,

$\int \frac{x4}{1 + x} dx$

Taking the constant term out,,,

$= > 4 \times \int \frac{x}{1 + x} dx$

By applying u-substitution,,

$= > \\ Integral \: Substitution \: Definition \\ \\ \\ \int f(g(x)) \times g(x)dx = \int f(u) \: du \: \: \: \: > \: \: Here,\: u = g(x)$
$= > u = 1 + x \\ \\ = > 1 + x - 1 = u - 1 \\ \\ Simplifying \\ \\ = > x = u - 1$

Therefore,,,

$= > 4 \times \int \frac{u - 1}{u} \: \: \: \frac{du}{dx}$

Substitute the value,,,

$= > u = 1 + x$

$4 \times \int \frac{u - 1}{u} \: \: \frac{d}{dx} (1 + x)$

By applying sum/difference rule ,,, we get,,,,

${(f +- \: g)}^{1} = f + - \: {g}^{1}$

$4 \times \int \frac{u - 1}{u} \frac{d}{dx} (1) + \frac{d}{dx} (x)$

Derivative of a constant is ,,,

$\frac{d}{dx} (a) = 0 \\ \\ \\ Therefore \\ \\ \\ = > 4 \times \int \frac{u - 1}{u} \frac{d}{dx} (x)$

Applying common derivative, that is,,,

$\frac{d}{dx} = 1$

Since,,,

$= > du = 1dx \\ \\ = > dx = 1du$

$= > 4 \times \int \frac{u - 1}{u} du$

Applying fraction rule,,, that is,,,

$\frac{a + - b}{c} = \frac{a}{c} + - \frac{b}{c} \\ \\ \\ = > 4 \times \int \frac{u}{u} - \frac{1}{u} du \\ \\ \\ = > 4 \times \int 1 - \frac{1}{u} du$

By applying the sum rule,,,,

$\int f(x) + - g(x)dx = \int f(x)dx + - g(x)$

$4 \times (\int 1du - \int \frac{1}{u} du)$

Since integral constant,,,

$\int adx = ax \\ \\ Therefore,,,, \\ \\ \\ 4 \times(\int 1 \times u - \int \frac{1}{u} du )$

By simplifying,,

$= > 4 \times (u - \int \frac{1}{u} du)$

By using the common integral,,,

$> \int \frac{1}{u} du = ln( |u| ) \\ \\ \\ = > 4 \times (u - ln( |u| ) )$

Substituting the value of,,,

$= > u = 1 + x$

Putting it back to the equation,,,,

$= > 4(1 + x - ln( |1+ x| ) )$

Adding a constant to our final solution,,,

$= > 4(1 + x - ln( |1 + x| ) ) + C$

Which is the required solution.

HOPE IT HELPS AND SOLVES YOUR DOUBTS ABOUT INTEGRALS!!!!!