Calculating the integral of a logarithmic expression. The problem I have been working with is int 1/(sqrt x(1+\sqrt x))dx The first step I did to solve this question was to set u=1+sqrt(x) the set du = (1/2) x^(-1/2) Then I set 1/2 ∫ln(du/u) and finally I think the answer is ln 1/x +c

Madison Costa

Madison Costa

Answered question

2022-11-02

Calculating the integral of a logarithmic expression.
The problem I have been working with is
1 x ( 1 + x ) d x
The first step I did to solve this question was to set u = 1 + x the set d u = ( 1 / 2 ) x 1 / 2
Then I set 1 2 ln ( d u / u ) and finally I think the answer is ln 1 x + c

Answer & Explanation

Jackson Trevino

Jackson Trevino

Beginner2022-11-03Added 14 answers

Setting u = 1 + x , we have
d u d x = 1 2 x d x x = 2 d u ,
so
1 1 + x d x x = 2 u d u = 2 ln u + C = 2 ln ( 1 + x ) + C .
Jared Lowe

Jared Lowe

Beginner2022-11-04Added 5 answers

u = x u 2 = x 2 u d u = d x d x x ( 1 + x ) = 2 u d u u ( 1 + u ) = 2 d u 1 + u etc.

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