Advertisements
Advertisements
प्रश्न
`int"x"^"n"."log" "x" "dx"`
योग
Advertisements
उत्तर
`int"x"^"n"."log" "x" "dx"`
= `int"log" "x" "x"^"n" "dx"`
`int"u"."v" "dx" = "u" int "v" "dx" - int ("du"/"dx") [int "v dx"] "dx"`
= `"log x" int "x"^"n" "dx" - int ["d"/"dx" ("log x")int "x"^"n" "dx"] "dx"`
= `"log x" xx ("x"^("n" + 1))/("n" + 1) - int 1/"x".("x"^("n" + 1))/("n" + 1) "dx"`
= `("x"^("n" + 1) "log x")/("n" + 1) - 1/("n" + 1) int"x"^"n" "dx"`
= `("x"^("n" + 1) "log x")/("n" + 1) - ("x"^("n" + 1))/("n" + 1)^2 + "C"`
= `("x"^("n" + 1) "log x")/("n" + 1) ["log x" - 1/("n" + 1)] + "C"`
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\sqrt{x}\left( x^3 - \frac{2}{x} \right) dx\]
\[\int \cos^{- 1} \left( \sin x \right) dx\]
\[\int\left( x + 2 \right) \sqrt{3x + 5} \text{dx} \]
\[\int\text{sin mx }\text{cos nx dx m }\neq n\]
\[\int\frac{1}{\sqrt{1 - \cos 2x}} dx\]
\[\int\frac{\log\left( 1 + \frac{1}{x} \right)}{x \left( 1 + x \right)} dx\]
\[\int\frac{\cos^3 x}{\sqrt{\sin x}} dx\]
\[\ \int\ x \left( 1 - x \right)^{23} dx\]
\[\int \sec^4 2x \text{ dx }\]
\[\int \cos^7 x \text{ dx } \]
\[\int x \cos^3 x^2 \sin x^2 \text{ dx }\]
\[\int\frac{1}{a^2 x^2 + b^2} dx\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 + 1}} dx\]
\[\int\frac{dx}{e^x + e^{- x}}\]
\[\int\frac{x}{\sqrt{4 - x^4}} dx\]
` ∫ {x-3} /{ x^2 + 2x - 4 } dx `
\[\int\frac{1}{2 + \sin x + \cos x} \text{ dx }\]
\[\int\frac{1}{3 + 4 \cot x} dx\]
\[\int e^\sqrt{x} \text{ dx }\]
\[\int\frac{\left( x \tan^{- 1} x \right)}{\left( 1 + x^2 \right)^{3/2}} \text{ dx }\]
∴\[\int e^{2x} \left( - \sin x + 2 \cos x \right) dx\]
\[\int\left( 2x - 5 \right) \sqrt{x^2 - 4x + 3} \text{ dx }\]
\[\int\frac{x^2 + 1}{\left( x - 2 \right)^2 \left( x + 3 \right)} dx\]
\[\int\frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{x^3 - 1}{x^3 + x} dx\]
\[\int\frac{1}{\left( 2 x^2 + 3 \right) \sqrt{x^2 - 4}} \text{ dx }\]
Write a value of
\[\int e^{3 \text{ log x}} x^4\text{ dx}\]
\[\int x^{\sin x} \left( \frac{\sin x}{x} + \cos x . \log x \right) dx\] is equal to
If \[\int\frac{\cos 8x + 1}{\tan 2x - \cot 2x} dx\]
\[\int \text{cosec}^2 x \text{ cos}^2 \text{ 2x dx} \]
\[\int\frac{1}{e^x + e^{- x}} dx\]
\[\int \cot^5 x\ dx\]
\[\int\frac{1}{4 x^2 + 4x + 5} dx\]
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
\[\int\frac{1}{1 - 2 \sin x} \text{ dx }\]
\[\int\sqrt{x^2 - a^2} \text{ dx}\]
\[\int x \sec^2 2x\ dx\]
\[\int \left( e^x + 1 \right)^2 e^x dx\]
\[\int \sin^3 \left( 2x + 1 \right) \text{dx}\]
