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基础部分
计算下列不定积分
(1) ∫ cos 3 x d x \int \cos^3 x d x ∫ cos 3 x d x ∫ sin 3 x d x \int \sin^3 x d x ∫ sin 3 x d x ∫ sec x d x \int \sec x d x ∫ sec x d x ∫ sec 3 x d x \int \sec^3 x d x ∫ sec 3 x d x ∫ 1 a 2 − x 2 d x ( a ≠ 0 ) \int \frac{1}{a^2-x^2} d x(a\neq 0) ∫ a 2 − x 2 1 d x ( a = 0 ) ∫ 1 x 2 − a 2 d x ( a ≠ 0 ) \int \frac{1}{x^2-a^2} d x(a\neq 0) ∫ x 2 − a 2 1 d x ( a = 0 ) ∫ 1 a 2 + x 2 d x ( a ≠ 0 ) \int \frac{1}{a^2+x^2} d x(a\neq 0) ∫ a 2 + x 2 1 d x ( a = 0 ) ∫ 1 a 2 + ( x + b ) 2 d x ( a ≠ 0 ) \int \frac{1}{a^2+(x+b)^2} d x(a\neq 0) ∫ a 2 + ( x + b ) 2 1 d x ( a = 0 ) ∫ 1 a 2 − ( x + b ) 2 d x ( a > 0 ) \int \frac{1}{a^2-(x+b)^2} d x(a>0) ∫ a 2 − ( x + b ) 2 1 d x ( a > 0 ) ∫ 1 ( x + b ) 2 − a 2 d x ( a > 0 ) \int \frac{1}{(x+b)^2-a^2} d x(a>0) ∫ ( x + b ) 2 − a 2 1 d x ( a > 0 ) ∫ 1 x 2 − a 2 d x ( a > 0 ) \int \frac{1}{\sqrt{x^2-a^2}} d x(a>0) ∫ x 2 − a 2 1 d x ( a > 0 ) ∫ 1 a 2 − x 2 d x ( a > 0 ) \int \frac{1}{\sqrt{a^2-x^2}} d x(a>0) ∫ a 2 − x 2 1 d x ( a > 0 ) ∫ 1 x 2 + a 2 d x ( a > 0 ) \int \frac{1}{\sqrt{x^2+a^2}} d x(a>0) ∫ x 2 + a 2 1 d x ( a > 0 ) ∫ csc 3 x d x \int \csc^3 x d x ∫ csc 3 x d x ∫ tan 2 x d x \int \tan^2 x d x ∫ tan 2 x d x ∫ tan 3 x d x \int \tan^3 x d x ∫ tan 3 x d x ∫ tan 4 x d x \int \tan^4 x d x ∫ tan 4 x d x ∫ cot 3 x d x \int \cot^3 x d x ∫ cot 3 x d x ∫ cos x 1 + sin x d x \int \frac{\cos x}{1+\sin x} d x ∫ 1 + s i n x c o s x d x ∫ 1 a 2 sin 2 x + b 2 cos 2 x d x \int \frac{1}{a^2 \sin^2 x+b^2 \cos^2 x} d x ∫ a 2 s i n 2 x + b 2 c o s 2 x 1 d x ∫ 1 sin 2 x d x \int \frac{1}{\sin 2x} d x ∫ s i n 2 x 1 d x ∫ 1 cos 2 x d x \int \frac{1}{\cos 2x} d x ∫ c o s 2 x 1 d x ∫ 1 a + b cos x d x ( a > 0 , b > 0 ) \int \frac{1}{a+b \cos x} d x(a>0,b>0) ∫ a + b c o s x 1 d x ( a > 0 , b > 0 ) ∫ 1 a + b sin x d x ( a > 0 , b > 0 ) \int \frac{1}{a+b \sin x} d x(a>0,b>0) ∫ a + b s i n x 1 d x ( a > 0 , b > 0 )
计算不定积分 ∫ e 2 x e x − 1 d x \int \frac{e^{2x}}{\sqrt{e^x-1}} d x ∫ e x − 1 e 2 x d x
计算不定积分 ∫ ln ( 1 + 1 + x x ) d x ( x > 0 ) \int \ln (1+\sqrt{\frac{1+x}{x}}) d x(x>0) ∫ ln ( 1 + x 1 + x ) d x ( x > 0 )
计算不定积分 ∫ e 2 x arctan e x − 1 d x \int e^{2x} \arctan \sqrt{e^x-1} d x ∫ e 2 x arctan e x − 1 d x
∫ 0 1 e − x d x = \int_{0}^{1} e^{-\sqrt{x}} d x= ∫ 0 1 e − x d x = 2 − 4 e 2-\frac{4}{e} 2 − e 4 1 − 2 e 1-\frac{2}{e} 1 − e 2 1 − 1 e 1-\frac{1}{e} 1 − e 1 ∫ 0 1 4 x − 3 x 2 − x + 1 d x = \int_{0}^{1} \frac{4x-3}{x^2-x+1} d x= ∫ 0 1 x 2 − x + 1 4 x − 3 d x = ∫ 0 1 arctan x x ( 1 + x ) d x = \int_{0}^{1} \frac{\arctan \sqrt{x}}{\sqrt{x}(1+x)} d x= ∫ 0 1 x ( 1 + x ) a r c t a n x d x = ∫ 0 π 4 sec 3 θ d θ = \int_{0}^{\frac{\pi}{4}} \sec^3 \theta d \theta= ∫ 0 4 π sec 3 θ d θ = 设连续函数 f ( x ) f(x) f ( x ) f ( x + 1 ) − f ( x ) = x ln x f(x+1)-f(x)=x \ln x f ( x + 1 ) − f ( x ) = x ln x ∫ 0 1 f ( x ) d x = 0 \int_{0}^{1} f(x) d x=0 ∫ 0 1 f ( x ) d x = 0 ∫ 1 2 f ( x ) d x = \int_{1}^{2} f(x) d x= ∫ 1 2 f ( x ) d x =
设 y = x 1 + x 2 y=\frac{x}{\sqrt{1+x^2}} y = 1 + x 2 x ∫ 1 2 3 2 x y d y = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} x y d y= ∫ 2 1 2 3 x y d y =
若 e − x e^{-x} e − x f ( x ) f(x) f ( x ) ∫ 1 2 1 x 2 f ( ln x ) d x = \int_{1}^{\sqrt{2}} \frac{1}{x^2} f(\ln x) d x= ∫ 1 2 x 2 1 f ( ln x ) d x = − 1 4 -\frac{1}{4} − 4 1 1 4 \frac{1}{4} 4 1
若函数 f ( x ) f(x) f ( x ) [ 0 , + ∞ ) [0,+\infty) [ 0 , + ∞ ) g ( x ) = ∫ 0 2 x f ( x + t 2 ) d t g(x)=\int_{0}^{2x} f(x+\frac{t}{2}) d t g ( x ) = ∫ 0 2 x f ( x + 2 t ) d t x → 0 + x\to 0^+ x → 0 + g ( x ) g(x) g ( x ) x \sqrt{x} x
设 f ( x ) f(x) f ( x ) x = 0 x=0 x = 0 x = 0 x=0 x = 0 f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0 φ ( x ) = { 1 x 2 ∫ 0 x t f ( t ) d t , x ≠ 0 0 , x = 0 \varphi(x)=\begin{cases}\frac{1}{x^2}\int_{0}^{x} t f(t) d t, & x\neq 0 \\ 0, & x=0\end{cases} φ ( x ) = { x 2 1 ∫ 0 x t f ( t ) d t , 0 , x = 0 x = 0 φ ( x ) \varphi(x) φ ( x ) x = 0 x=0 x = 0 φ ′ ( x ) \varphi'(x) φ ′ ( x ) x = 0 x=0 x = 0 φ ′ ( x ) \varphi'(x) φ ′ ( x ) x = 0 x=0 x = 0
若连续周期函数 y = f ( x ) y=f(x) y = f ( x ) x x x f ( x + 3 ) + f ( x − 3 ) = f ( x ) f(x+3)+f(x-3)=f(x) f ( x + 3 ) + f ( x − 3 ) = f ( x ) f ( x ) f(x) f ( x )
设 f ( x ) f(x) f ( x ) [ − a , a ] [-a,a] [ − a , a ] a > 0 a>0 a > 0 g ( x ) = ∫ − a a ∣ x − t ∣ f ( t ) d t g(x)=\int_{-a}^{a}|x-t| f(t) d t g ( x ) = ∫ − a a ∣ x − t ∣ f ( t ) d t [ − a , a ] [-a,a] [ − a , a ] g ( x ) g(x) g ( x ) g ( x ) g(x) g ( x ) g ( x ) g(x) g ( x ) g ( x ) g(x) g ( x )
若 F ( x ) = ∫ − π π ∣ x − t ∣ sin t d t F(x)=\int_{-\pi}^{\pi}|x-t| \sin t d t F ( x ) = ∫ − π π ∣ x − t ∣ sin t d t F ′ ( 0 ) = F'(0)= F ′ ( 0 ) =
若函数 y ( x ) = ∫ 2 x 2 e − t d t y(x)=\int_{2}^{x^2} e^{-\sqrt{t}} d t y ( x ) = ∫ 2 x 2 e − t d t d 2 [ y ( x ) ] d x 2 ∣ x = − 1 = \frac{d^2[y(x)]}{d x^2}|_{x=-1}= d x 2 d 2 [ y ( x )] ∣ x = − 1 = 4 e − 1 4e^{-1} 4 e − 1 4 e 4e 4 e
已知函数 f ( x ) = ∫ 1 x 1 + t 3 d t f(x)=\int_{1}^{x} \sqrt{1+t^3} d t f ( x ) = ∫ 1 x 1 + t 3 d t ∫ 0 1 x f ( x ) d x = \int_{0}^{1} x f(x) d x= ∫ 0 1 x f ( x ) d x =
设连续函数 f ( x ) f(x) f ( x ) ∫ 0 x f ( t ) d t = x e x \int_{0}^{x} f(t) d t=x e^x ∫ 0 x f ( t ) d t = x e x ∫ 1 e f ( ln x ) x d x = \int_{1}^{e} \frac{f(\ln x)}{x} d x= ∫ 1 e x f ( l n x ) d x =
设 a n = ∫ 0 1 x n 1 − x 2 d x ( n = 0 , 1 , 2 , ⋯ ) a_n=\int_{0}^{1} x^n \sqrt{1-x^2} d x(n=0,1,2,\cdots) a n = ∫ 0 1 x n 1 − x 2 d x ( n = 0 , 1 , 2 , ⋯ ) lim n → ∞ ( a n a n − 2 ) n = \lim _{n \to \infty}(\frac{a_n}{a_{n-2}})^n= lim n → ∞ ( a n − 2 a n ) n =
∫ 5 5 1 ∣ x 2 − 9 ∣ d x = \int_{\sqrt{5}}^{5} \frac{1}{\sqrt{|x^2-9|}} d x= ∫ 5 5 ∣ x 2 − 9∣ 1 d x = ∫ − ∞ + ∞ ∣ x ∣ e − x 2 d x = \int_{-\infty}^{+\infty}|x| e^{-x^2} d x= ∫ − ∞ + ∞ ∣ x ∣ e − x 2 d x = 设 f ( x ) = lim n → ∞ 1 − x 2 n 1 + x 2 n x f(x)=\lim _{n \to \infty} \frac{1-x^{2n}}{1+x^{2n}} x f ( x ) = lim n → ∞ 1 + x 2 n 1 − x 2 n x ∫ 0 2 f ( x ) d x = a \int_{0}^{2} f(x) d x=a ∫ 0 2 f ( x ) d x = a a = a= a =
∫ − 1 1 ( 1 1 + 2 − 1 x ) ′ d x = \int_{-1}^{1}\left(\frac{1}{1+2^{-\frac{1}{x}}}\right)' d x= ∫ − 1 1 ( 1 + 2 − x 1 1 ) ′ d x = ∫ − 1 2 1 2 ∣ x ∣ ( arcsin x + arccos x ) 1 − x 2 d x = \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{|x|(\arcsin x+\arccos x)}{\sqrt{1-x^2}} d x= ∫ − 2 1 2 1 1 − x 2 ∣ x ∣ ( a r c s i n x + a r c c o s x ) d x = ∫ 1 2 1 ( 1 − x ) arcsin ( 1 − x ) 2 x − x 2 d x = \int_{\frac{1}{2}}^{1} \frac{(1-x) \arcsin (1-x)}{\sqrt{2x-x^2}} d x= ∫ 2 1 1 2 x − x 2 ( 1 − x ) a r c s i n ( 1 − x ) d x = 若 f ( x ) = 1 1 + x 2 + x 3 ∫ 0 1 f ( x ) d x f(x)=\frac{1}{1+x^2}+x^3 \int_{0}^{1} f(x) d x f ( x ) = 1 + x 2 1 + x 3 ∫ 0 1 f ( x ) d x f ( x ) = f(x)= f ( x ) =
∫ − 1 1 [ x 3 cos x + ln ( x + x 2 + 1 ) ] d x = \int_{-1}^{1}\left[x^3 \cos x+\ln \left(x+\sqrt{x^2+1}\right)\right] d x= ∫ − 1 1 [ x 3 cos x + ln ( x + x 2 + 1 ) ] d x = 设 f ( x ) = ∫ 0 x cos t 1 + sin 2 t d t f(x)=\int_{0}^{x} \frac{\cos t}{1+\sin^2 t} d t f ( x ) = ∫ 0 x 1 + s i n 2 t c o s t d t ∫ 0 π 2 f ′ ( x ) 1 + f 2 ( x ) d x = \int_{0}^{\frac{\pi}{2}} \frac{f'(x)}{1+f^2(x)} d x= ∫ 0 2 π 1 + f 2 ( x ) f ′ ( x ) d x = − π 4 -\frac{\pi}{4} − 4 π − arctan π 4 -\arctan \frac{\pi}{4} − arctan 4 π π 4 \frac{\pi}{4} 4 π arctan π 4 \arctan \frac{\pi}{4} arctan 4 π
设 ∫ 1 + ∞ f ( x ) d x = A \int_{1}^{+\infty} f(x) d x=A ∫ 1 + ∞ f ( x ) d x = A f ( x ) = x 3 e − x 2 + 1 x ( 1 + x ) A f(x)=x^3 e^{-x^2}+\frac{1}{x(1+x)} A f ( x ) = x 3 e − x 2 + x ( 1 + x ) 1 A ∫ 1 + ∞ f ( x ) d x = \int_{1}^{+\infty} f(x) d x= ∫ 1 + ∞ f ( x ) d x = 1 1 − ln 2 \frac{1}{1-\ln 2} 1 − l n 2 1 1 ln 2 \frac{1}{\ln 2} l n 2 1 e 1 − ln 2 \frac{e}{1-\ln 2} 1 − l n 2 e 1 ( 1 − ln 2 ) e \frac{1}{(1-\ln 2)e} ( 1 − l n 2 ) e 1
函数 f ( x ) = { 1 1 + x 2 , x ≤ 0 ( x + 1 ) e x ( x + 2 ) 2 , x > 0 f(x)=\begin{cases}\frac{1}{\sqrt{1+x^2}}, & x\leq 0 \\ \frac{(x+1)e^x}{(x+2)^2}, & x>0\end{cases} f ( x ) = { 1 + x 2 1 , ( x + 2 ) 2 ( x + 1 ) e x , x ≤ 0 x > 0 F ( x ) = { ln ( 1 + x 2 − x ) , x ≤ 0 ( x − 1 ) e x , x > 0 F(x)=\begin{cases}\ln \left(\sqrt{1+x^2}-x\right), & x\leq 0 \\ (x-1)e^x, & x>0\end{cases} F ( x ) = { ln ( 1 + x 2 − x ) , ( x − 1 ) e x , x ≤ 0 x > 0 F ( x ) = { ln ( 1 + x 2 − x ) + 1 , x ≤ 0 ( x − 1 ) e x , x > 0 F(x)=\begin{cases}\ln \left(\sqrt{1+x^2}-x\right)+1, & x\leq 0 \\ (x-1)e^x, & x>0\end{cases} F ( x ) = { ln ( 1 + x 2 − x ) + 1 , ( x − 1 ) e x , x ≤ 0 x > 0 F ( x ) = { ln ( 1 + x 2 + x ) + 1 , x ≤ 0 x e x , x > 0 F(x)=\begin{cases}\ln \left(\sqrt{1+x^2}+x\right)+1, & x\leq 0 \\ x e^x, & x>0\end{cases} F ( x ) = { ln ( 1 + x 2 + x ) + 1 , x e x , x ≤ 0 x > 0 F ( x ) = { ln ( 1 + x 2 + x ) + 1 , x ≤ 0 x e x + 1 , x > 0 F(x)=\begin{cases}\ln \left(\sqrt{1+x^2}+x\right)+1, & x\leq 0 \\ x e^x+1, & x>0\end{cases} F ( x ) = { ln ( 1 + x 2 + x ) + 1 , x e x + 1 , x ≤ 0 x > 0
∫ 0 1 2 x arcsin x 1 − x 2 d x = \int_{0}^{\frac{1}{2}} \frac{x \arcsin x}{\sqrt{1-x^2}} d x= ∫ 0 2 1 1 − x 2 x a r c s i n x d x = ∫ 0 1 x arcsin ( 1 − x ) d x = \int_{0}^{1} x \arcsin (1-x) d x= ∫ 0 1 x arcsin ( 1 − x ) d x = 设 y = y ( x ) y=y(x) y = y ( x ) x y ′ = 1 − x 2 x y'=\sqrt{1-x^2} x y ′ = 1 − x 2 y ( 1 ) = 0 y(1)=0 y ( 1 ) = 0 ∫ 0 1 y ( x ) d x = \int_{0}^{1} y(x) d x= ∫ 0 1 y ( x ) d x =
∫ 1 + ∞ 1 x ( x + 2 ) d x = \int_{1}^{+\infty} \frac{1}{x(x+2)} d x= ∫ 1 + ∞ x ( x + 2 ) 1 d x = 设 f ′ ( e x ) = sin x f'(e^x)=\sin x f ′ ( e x ) = sin x f ( x ) f(x) f ( x )
强化部分
∫ 0 1 ln 1 1 − x d x = \int_{0}^{1} \ln \frac{1}{1-x} d x= ∫ 0 1 ln 1 − x 1 d x = 求 ∫ x + 2 ( 2 x + 1 ) ( x 2 + x + 1 ) d x \int \frac{x+2}{(2x+1)(x^2+x+1)} d x ∫ ( 2 x + 1 ) ( x 2 + x + 1 ) x + 2 d x
∫ 1 e cos ( ln x ) d x = \int_{1}^{e} \cos (\ln x) d x= ∫ 1 e cos ( ln x ) d x = 设函数 f ( x ) f(x) f ( x ) x f ( x ) + f ( 1 − x ) = x 2 x f(x)+f(1-x)=x^2 x f ( x ) + f ( 1 − x ) = x 2 ∫ f ( x ) d x \int f(x) d x ∫ f ( x ) d x
∑ n = 1 ∞ ∫ n n + 1 2 − x d x = \sum_{n=1}^{\infty} \int_{n}^{n+1} 2^{-\sqrt{x}} d x= ∑ n = 1 ∞ ∫ n n + 1 2 − x d x = 已知 f ( x ) f(x) f ( x ) ∫ 0 1 f ( x ) d x = 2 \int_{0}^{1} f(x) d x=2 ∫ 0 1 f ( x ) d x = 2 ∫ 0 2 x f ( 1 − x ) d x = \int_{0}^{2} x f(1-x) d x= ∫ 0 2 x f ( 1 − x ) d x =
已知 f ( x ) f(x) f ( x ) f ( x 2 + 1 ) − f ( x 2 ) = x ( x > 0 ) f(x^2+1)-f(x^2)=x(x>0) f ( x 2 + 1 ) − f ( x 2 ) = x ( x > 0 ) ∫ 0 1 f ( x ) d x = 1 \int_{0}^{1} f(x) d x=1 ∫ 0 1 f ( x ) d x = 1 ∫ 1 2 f ( x ) d x = \int_{1}^{2} f(x) d x= ∫ 1 2 f ( x ) d x =
设 f ( t ) = ∫ 0 1 t ∣ t − x ∣ d x f(t)=\int_{0}^{1} t|t-x| d x f ( t ) = ∫ 0 1 t ∣ t − x ∣ d x ∫ − 1 2 f ( t ) d t \int_{-1}^{2} f(t) d t ∫ − 1 2 f ( t ) d t
设 f ( x ) f(x) f ( x ) ∫ 0 2 f ( x ) d x = 1 \int_{0}^{2} f(x) d x=1 ∫ 0 2 f ( x ) d x = 1 g ( x ) = { x , x > 0 1 x − 2 , x < 0 g(x)=\begin{cases}x, & x>0 \\ \frac{1}{x}-2, & x<0\end{cases} g ( x ) = { x , x 1 − 2 , x > 0 x < 0 ∫ 0 2 f [ g ( x ) ] d x = \int_{0}^{2} f[g(x)] d x= ∫ 0 2 f [ g ( x )] d x =
已知 f ′ ( x ) = arctan ( x − 1 ) 2 f'(x)=\arctan (x-1)^2 f ′ ( x ) = arctan ( x − 1 ) 2 f ( 0 ) = 0 f(0)=0 f ( 0 ) = 0 ∫ 0 1 f ( x ) d x = \int_{0}^{1} f(x) d x= ∫ 0 1 f ( x ) d x =
设 g ( x ) = x 2 g(x)=x^2 g ( x ) = x 2 g [ f ( x ) ] = − x 2 + 2 x + 3 g[f(x)]=-x^2+2x+3 g [ f ( x )] = − x 2 + 2 x + 3 f ( x ) > 0 f(x)>0 f ( x ) > 0 ∫ 0 1 1 f ( x ) d x = \int_{0}^{1} \frac{1}{f(x)} d x= ∫ 0 1 f ( x ) 1 d x =
求 ∫ − 1 1 x ln ( 1 + e x ) d x \int_{-1}^{1} x \ln \left(1+e^x\right) d x ∫ − 1 1 x ln ( 1 + e x ) d x
求 ∫ 0 1 d x ( x + 1 ) ( x 2 + 1 ) \int_{0}^{1} \frac{d x}{(x+1)(x^2+1)} ∫ 0 1 ( x + 1 ) ( x 2 + 1 ) d x
设 F ( x ) > 0 F(x)>0 F ( x ) > 0 R \mathbb{R} R F ( 0 ) = π F(0)=\sqrt{\pi} F ( 0 ) = π F ( x ) F ′ ( x ) = cos x 2 sin 2 x + cos 2 x F(x) F'(x)=\frac{\cos x}{2 \sin^2 x+\cos^2 x} F ( x ) F ′ ( x ) = 2 s i n 2 x + c o s 2 x c o s x F ( x ) F(x) F ( x )
求 ∫ 0 1 arcsin x x ( 1 − x ) d x \int_{0}^{1} \frac{\arcsin \sqrt{x}}{\sqrt{x(1-x)}} d x ∫ 0 1 x ( 1 − x ) a r c s i n x d x
求 ∫ 0 1 4 d x sin 2 x + 3 cos 2 x \int_{0}^{\frac{1}{4}} \frac{d x}{\sin^2 x+3 \cos^2 x} ∫ 0 4 1 s i n 2 x + 3 c o s 2 x d x
求 ∫ 1 3 d x ∣ 2 x − x 2 ∣ \int_{1}^{3} \frac{d x}{\sqrt{|2x-x^2|}} ∫ 1 3 ∣2 x − x 2 ∣ d x
设 n n n ∫ 0 1 x 2 ln n x d x = \int_{0}^{1} x^2 \ln^n x d x= ∫ 0 1 x 2 ln n x d x =
设 f ( x ) = ∫ 0 x sin t t d t , 0 ≤ x ≤ 1 f(x)=\int_{0}^{x} \frac{\sin t}{\sqrt{t}} d t,0\leq x\leq 1 f ( x ) = ∫ 0 x t s i n t d t , 0 ≤ x ≤ 1 f + ′ ( 0 ) = f_{+}'(0)= f + ′ ( 0 ) = − π 2 -\frac{\pi}{2} − 2 π π 2 \frac{\pi}{2} 2 π − π -\pi − π π \pi π
设 ∣ x ∣ ≤ 1 |x|\leq 1 ∣ x ∣ ≤ 1 I ( x ) = ∫ − 1 1 ∣ t − x ∣ e 2 t d t I(x)=\int_{-1}^{1}|t-x| e^{2t} d t I ( x ) = ∫ − 1 1 ∣ t − x ∣ e 2 t d t
设函数 f ( x ) = ∫ 0 1 ∣ t 2 − x 2 ∣ d t ( x > 0 ) f(x)=\int_{0}^{1}|t^2-x^2| d t(x>0) f ( x ) = ∫ 0 1 ∣ t 2 − x 2 ∣ d t ( x > 0 ) f ′ ( x ) f'(x) f ′ ( x ) f ( x ) f(x) f ( x )
设 f ( x ) = x , g ( x ) = { cos x , x ≤ π 0 , x > π f(x)=x,g(x)=\begin{cases}\cos x, & x\leq\pi \\ 0, & x>\pi\end{cases} f ( x ) = x , g ( x ) = { cos x , 0 , x ≤ π x > π F ( x ) = ∫ 0 x f ( t ) g ( x − t ) d t ( x ≥ 0 ) F(x)=\int_{0}^{x} f(t) g(x-t) d t(x\geq 0) F ( x ) = ∫ 0 x f ( t ) g ( x − t ) d t ( x ≥ 0 )
设 y = f ( x ) = x ∫ 0 2 e − ( x t ) 2 d t + x 2 y=f(x)=x \int_{0}^{2} e^{-(x t)^2} d t+x^2 y = f ( x ) = x ∫ 0 2 e − ( x t ) 2 d t + x 2 x = 0 x=0 x = 0 x = g ( y ) x=g(y) x = g ( y ) g ′ ′ ( 0 ) = g''(0)= g ′′ ( 0 ) =
F ( x ) = ∫ 0 π 2 ∣ sin x − sin t ∣ d t ( x ≥ 0 ) F(x)=\int_{0}^{\frac{\pi}{2}}|\sin x-\sin t| d t(x\geq 0) F ( x ) = ∫ 0 2 π ∣ sin x − sin t ∣ d t ( x ≥ 0 ) x → 0 + x\to 0^+ x → 0 + a + b x + c x 2 a+b x+c x^2 a + b x + c x 2 a , b , c a,b,c a , b , c 求 ∫ 1 2 d x x 3 x 2 − 2 x − 1 \int_{1}^{2} \frac{d x}{x \sqrt{3x^2-2x-1}} ∫ 1 2 x 3 x 2 − 2 x − 1 d x
∫ − 1 5 1 1 + 2 x − 2 3 d x = \int_{-1}^{5} \frac{1}{1+2^{\sqrt[3]{x-2}}} d x= ∫ − 1 5 1 + 2 3 x − 2 1 d x = 3 \sqrt{3} 3 3 2 \frac{3}{2} 2 3 设 y = y ( x ) y=y(x) y = y ( x ) x 2 y ′ + ( x 2 − 3 ) y 2 = 0 x^2 y'+(x^2-3) y^2=0 x 2 y ′ + ( x 2 − 3 ) y 2 = 0 y ( 1 ) = 1 y(1)=1 y ( 1 ) = 1 y = y ( x ) y=y(x) y = y ( x ) ∫ 0 3 y 2 ( x ) d x \int_{0}^{3} y^2(x) d x ∫ 0 3 y 2 ( x ) d x
∫ 0 + ∞ x ln x 1 + x 4 d x = \int_{0}^{+\infty} \frac{x \ln x}{1+x^4} d x= ∫ 0 + ∞ 1 + x 4 x l n x d x =