Question

integaral of log (1+x)​

Answer 1

Answer:

thank you for the free points mate

Answer 2

$\large\underline\blue{\bold{Given\:Question - }}$

$\tt \longmapsto\: \int \: log(1 + x)dx$

$\begin{gathered}\Large{\bold{\red{\underline{Formula \: Used \::}}}} \end{gathered}$

$(1). \: \boxed{ \green{ \bf \: \int1 \: dx \: = x \: + \: c}}$

$(2). \:\boxed{ \green{ \bf \: \int\dfrac{1}{x}dx = log(x) + c }}$

$(3). \:\boxed{ \green{ \bf \: \dfrac{d}{dx} log(x) = \dfrac{1}{x} }}$

Concept Used :-

Integration by Parts

• Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways.

✏️See the rule:

• ∫u v dx = u∫v dx −∫u' (∫v dx) dx

• u is the function u(x)

• v is the function v(x)

• u' is the derivative of the function u(x)

For integration by parts , the ILATE rule is used to choose u and v.

where,

• I - Inverse trigonometric functions

• L -Logarithmic functions

• A - Arithmetic and Algebraic functions

• T - Trigonometric functions

• E- Exponential functions

The alphabet which comes first is choosen as u and other as v.

Let's do it now!!

$\large\underline\purple{\bold{Solution :- }}$

$\tt \longmapsto\: \int \: log(1 + x)dx$

$\tt \longmapsto\: \int \: 1. log(1 + x)dx$

Here,

• u = log(1 + x)

• v = 1

So,

• By using Integration by Parts, we get

$\tt \longmapsto\: log(1 + x) \int1dx - \int \{\dfrac{d}{dx} log(1 + x) \int1dx \}dx$

$\tt \longmapsto\:x log(1 + x) - \int\dfrac{1}{1 + x} \times x \: dx$

$\tt \longmapsto\:x log(1 + x) - \int\dfrac{x}{1 + x} \: dx$

$\tt \longmapsto\:x log(1 + x) - \int\dfrac{x + 1 - 1}{1 + x} \: dx$

$\tt \longmapsto\:x log(1 + x) - \int(\dfrac{1 + x}{1 + x} - \dfrac{1}{x + 1} )\: dx$

$\tt \longmapsto\:x log(1 + x) - \int(1 - \dfrac{1}{1 + x})\: dx$

$\tt \longmapsto\:x log(1 + x) - x + log(1 + x) + c$

$\tt \longmapsto\:(x + 1) \: log(1 + x) - x + c$