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) ...
【解析】 $$ 2 \ln 2 - 1 $$ 结果一 题目 【题目】用Newton-Leibniz公式计算下列定积分∫_(-1)^1|x|dx 答案 【解析】∫_(-1)^1|x|dx=∫_(-1)^9-xdx+∫_0^1xdx=-1/2x^2|_(-1)^0+1/2x^2 或∫_(-1)^1|x|dx=2∫_0^1|x|dx=2∫_0^1xdx=x^2∫_0^1=1.(...
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 ...
Integrate: \int x^n \ln x dx Integrate: \int (x^2-7)^5(2x) dx Integrate: \int x \sqrt {2x + 1} dx Integrate \int x \tan^{-1} x dx Integrate \int \frac{\sqrt{25-x^2{x^2} \, dx Integrate: \int \frac{1}{(x)\ln(x)} dx ...
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 ...
【解析】 解:由定积分分部积分公式 $$ \int _ { 1 } ^ { e } x \ln x d x = \int _ { 1 } ^ { e } \ln x d \frac { x ^ { 2 } } { 2 } - \frac { x ^ { 2 } } { 2 } \ln x | _ { 1 } ^ { e } - \int _ { 1 } ^ { e } \frac { x ^ { ...
解析 【解析】 $$,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) ...
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: ...
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...
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