解析 【解析】 解 因为 $$ \frac { 1 } { x } $$dx凑微分为dlnx,故 $$ \int \frac { \ln x } { x } d x = \int \ln x d \ln x = \frac { 1 } { 2 } \ln ^ { 2 } x + C $$ 结果一 题目 35cm灬24cm 答案 【解析】|a| 结果二 题目 【题目】五、看图,补全单词...
Find dy if y = 3 ln (4 + x^4). Find d/dx ln (1 + x). Find y' and y'' for y = ln x/(1 + ln x). Find the value of \ln 2^{1000} if \ln 2 = 0.69314. Evaluate int_1^infty ln x x^2 dx. Evaluate lim x ln x x ...
【解析】 $$ \int x \ln ( x - 1 ) d x = \int \ln ( x - 1 ) d ( \frac { x ^ { 2 } } { 2 } ) \\ = \frac { x ^ { 2 } } { 2 } \ln ( x - 1 ) - \int \frac { x ^ { 2 } } { 2 ( x - 1 ) } d x \\ = \frac { x ^ { 2 }...
View Solution ∫(x−1)dx(1+x)(1+x2) View Solution ∫ln(√1+x+√1−x)dx View Solution Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation ...
Evaluate ∫ln(1+x)dx. Integration by Parts:Integration by parts is an integration technique implemented to integrate a product of two functions. To apply integration by parts to evaluate an integral of a product, simply follow the formula given below: ...
解析 $$,解: \int1n(1+x)dx \\ =x \ln(1+x)- \int \frac{x}{1+x}dx \\ =x \ln(1+x)- \int \frac{1+x-1}{1+x}dx \\ =x \ln(1+x)- \int 1dx+ \int \frac{1}{1+x}dx \\ =x \ln(1+x)-x+1n(1+x)+c(c为代数 $$ 倍常数) ...
∫f(x)⋅g(x)dx=f(x)∫g(x)dx−∫[∫g(x)dxd(f(x))dx]dx. Answer and Explanation:1 We are given integral ∫xlnxdx Perform the indefinite integral using integration by parts: {eq}\begin{align} \implies \int x \ln... ...
解析 【解析】 $$,解: \int1n(1+x)dx \\ =x \ln(1+x)- \int \frac{x}{1+x}dx \\ =x \ln(1+x)- \int \frac{1+x-1}{1+x}dx \\ =x \ln(1+x)- \int 1dx+ \int \frac{1}{1+x}dx \\ =x \ln(1+x)-x+1n(1+x)+c $$ (c为任意常数) ...
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Integrate: {eq}\int x^n \ln x dx{/eq} Method of Integration by Parts: Integration by parts is a special method of integration that is very often useful when two integrable functions are multiplied together. If we suppose that {eq}f {/eq} and {eq}g {/eq} are two real valued fun...